This is not going to be the answer you were hoping for, unfortunately.
The term 'phosphor' refers to any material that exhibits some form of luminescence. The key word there is some form. There are many varieties, and specific phosphors exhibit a specific kind of luminescence.
They cannot be used interchangeably. Your neon tube phosphors are fluorescent, which means they will emit light of a different frequency (color) if the right light source is shined on it. In other words, and I am sure you're very familiar with this, argon and mercury vapor produces ultraviolet light, which hits the phosphor coating, causing the phosphor to glow in one of hundreds of possible colors available in neon lightning.
That is the only way those phosphors can be made to emit light. They are fluorescent and nothing else. They are not phosphorescent (continues to glow after the light source is removed), nor are they chemiluminescent. And, sadly, they are certainly not electroluminescent.
In other words, all of your neon sign phosphors are completely useless in the context of EL wire. None of them can be made to emit light via electroluminescence. Sorry.
You can only make EL wire with electroluminescent phosphors, which are all semiconducting. And there is a grand total of 3 that are of any practical use (i.e., not so horribly inefficient that producing usable amounts of light would quickly overheat the display) for EL lighting.
Those three electroluminescent phosphors are all zinc sulfide, doped with either copper (greenish), silver (blue), or manganese (yellow-orange). Those are the only 3 things you can use. Because those are specifically electroluminescent. Electroluminescence is a fairly rare property.
However, there is a different way your phosphors can be (and probably already are) used with EL wire.
The different colors in EL wire are done in exactly the same way as your neon lamps, albeit with a different light source. Blue-green light emitted by the EL phosphor core of the wire, which makes a fluorescent dye in an outer PVC jacket around the wire fluoresce a specific and different color. So, while you must use zinc sulfide doped with copper, silver or manganese to produce the light, you can certainly do whatever you wish with this light once it is emitted. If you have neon phosphors that respond well to blue or green light, then you can definitely use them as a coating for the EL wire, though it will need to be added as part of the PVC manufacturing process to make anything remotely robust. Soda paste won't cut it, as it is on the outside, instead of coating the inside of a glass tube. And it must be flexible.
I would wager that some (if not all) of the fluorescent phosphors used in the PVC jackets around EL wire are in fact the exact phosphors you use to make those same colors in neon lamps. Only mixed in with clear plastic, rather than coating the inside of a glass tube.
It might be worth using a neon tube with EL wire inside instead of gas, and using that wire as the light source for the phosphor, only painted inside the tube like any other neon light. It would be dim unless you used a bundle of EL wires, but could produce a cool effect.
I know that is not what you were imagining or hoping, but thats the cards physics has given us, sadly.
The reasoning behind wire sizing is to provide less than allowable voltage drop over the distance between power source and destination.
You are doing all your calculations without understanding this goal, jumping between clearly defined terms as voltage drop, and undefined terms as conductance of unspecified copper alloy.
Start with a clear goal: voltage drop must be less than 3%. At 48 V this gives 1.44V, which you skipped to calculate and post.
Now, you have 2 wires, hot wire, and return wire. To get less than 1.44V at the load, you need to divide this value by two, since both wires will have the voltage drop. Which gives 0.72V per wire, assuming the wires are identical.
This gives you the limit on how much wire resistance you can afford at 3A load: 0.72V/3A = 0.24 Ohms. If your distance is 10m, then the wire must have 0.024 Ohm/m, a clearly identifiable parameter. This concludes the first part of the task.
Now which wire to select for this job, is up to unspecified condition in your exam task - it could be a copper-based alloy wire, it could be aluminum wire, or could be copper-coated steel wire. The rated carrying capacity will also depend on wire insulation and requirements for overheating. So it is a plus-minus guardbands set by manufacturer.
The first calculator is goofy because it does not count for allowable voltage drop, and its output is nonsense. The second calculator gives 16AWG copper for 2% loss (they don't have 1.5% entry), which looks reasonable.
So, what is the question again?
Best Answer
Caveat: I've not built EL wire, but having read some articles and having the background to analyse this here are some guide lines.
How it works: An electric field kicks electrons up to higher orbitals within the phosphor molecule which when then decay/emit light. This orbital decay and light emission will be very fast (called prompt emission)but we'd need to know the phosphor characteristics to be precise. But certainly ns to us range is reasonable. That means that for every AC cycle you get a pulse of light, if you have higher numbers of cycles in a second you get more pulses of light and therefore it is brighter.
The higher voltage effect is a little harder to understand. Strictly speaking, once you've ionized the phosphor the additional voltage will go into heat for a given electron in it's orbital. But depending upon the particular phosphor, if it has higher vacancies available then the electron might skip to a higher energy and then emit a subsequently higher energy photon (bluer in color). This is again dependant upon formulation of the phosphor. So if you look at an individual molecule you can't explain why higher voltage gives you higher light output. I strongly suspect that it has to do with the fact that the electric field around the thin wire excites more of a volume of phosphor at higher voltages. As the voltage increases, the e-field lines move further out, which then interacts with more phosphor which means that there is greater light.
With this in mind how do you increase output?
Make your windings closer together, you have a trade off with the wire blocking more light and the E-fields being more strongly overlapping and exciting more volume of phosphor. Commercial EL wires use a transparent electrode that wraps around the phosphor eliminating this effect (i.e. uniform E-field).
Jeri coats the wire on the outside with phosphor, the E-Field is lower outside of the wire, so if you can coat the whole assembly with transparent HIGH dielectric insulating material, more of the E-Field will exist on the outside where the phosphor is.
Get a core wire with an insulator that has LOWER dielectric constant, this will cause the E-Field on the outside to increase. If the dielectric constant of the core wire is already low then there isn't too much you can do.
Try different configurations: remember that E-Filed is Voltage divided by distance so a lower voltage as a tighter spacing will have a high E-field. I can think of several variations. Two thinner wires wound around each other, 4 very thin wires would around each other with alternate pairs connected together.
Safety note: if you are building the coaxial design (as per the video) drive the inner electrode with the HV, it will be shielded just that much more.