Ditch R1. Yes, 0.001\$\Omega\$ resistors exist, but what would you do with it? At 2A it will drop 2mV. The collector current is defined by the base current, there's no need to limit it this way (if that would have been the purpose).
You don't necessarily need a MOSFET to switch 2A, but if you use a BJT it will probably have to be a Darlington. OTOH a MOSFET is much faster than a BJT, so better suited for PWM work.
Then the flyback diode. It's wrong polarized, but you mention that in the question, so I won't say anything about that.
If the solenoid is a 24V/2A type its resistance will be 12\$\Omega\$, not 5.6.
I missed the line that says you'll be driving it from a microcontroller. The following assumes you drive it from 24V, like in the schematic. Later I'll make a note about the microcontroller.
Then R2. Assuming a solenoid of 12\$\Omega\$, and an \$H_{FE}\$ for your Darlington of 100, then from the base this will look as a 1200\$\Omega\$ resistance. You'll need 20mA of base current. With a voltage of 22V (24V minus a couple BE junctions) that would mean you should have maximum 1100\$\Omega\$ for R2 + \$H_{FE}\$ \$\times\$R3. So even without R2 you won't get the 2A. You'll need a transistor with a higher \$H_{FE}\$.
But even then R2 won't be necessary. With an \$H_{FE}\$ of 1000, if the base current would be higher than 2mA the transistor will saturate and the solenoid will limit the collector current to 2A.
Important notice on the common collector configuration you're using. Even if you would drive it from 24V the emitter voltage won't be 24V, but 22V. The base voltage will be 24V maximum, and if it would drive the emitter higher than 24V minus 2 BE junctions there wouldn't be no current anymore.
If you drive it from a 5V microcontroller the emitter voltage won't go higher than 3V! Again, if it would be higher there wouldn't flow any base current. You might use a common collector with a 24V input, but not with 5V.
Usually you'll use a common emitter configuration, where the solenoid comes at the place of R1. In that case you'll need R2. If your microcontroller runs at 5V and you're using the KSD1222 (see below), you'll have a voltage drop of 5V - 2V = 3V across R2. You'll need at least 2mA, but let's play safe and give it 10mA. Then R2 should be maximum 3V/10mA = 300\$\Omega\$.
If you want to use a MOSFET the Si2318DS is suitable. It's a 40V FET which can drive 3A at less than 4V \$V_{GS}\$. \$R_{DS(ON)}\$ is 45m\$\Omega\$, so at 2A it will only dissipate 180mW. That sounds safe, but when you're going to PWM this will rise due to switching losses. At 300Hz this will not really be a problem, however.
If you would want to use the Darlington, the KSD1222 is also a 40V type, with \$H_{FE}\$ of minimum 1000. Can drive 3A. But here saturation voltage can be as high as 1.5V. At 2A this means the transistor will dissipate 3W, so you'll need a heatsink. The MOSFET is the better solution.
This is a difficult problem to cover in a couple of hundred words, so this will be brief and you'll just have to do some research on your own. But I'll try to summarize it enough so you at least know what to research.
You need to know about trace impedance, signal termination, signal return paths, and bypass/decoupling caps. If you got these absolutely correct then you would have zero EMC problems. Getting it 100% perfect is impossible, but you can get much closer than you are now.
First, let's look at signal return paths... For every signal there must be a return path. Normally the return is on the power or ground plane, but it could be somewhere else too. On your PCB, the return is on a plane. The return path goes from the receiver back to the driver. The loop area is the physical loop created by the signal plus the return path. Normally the laws of physics will cause the loop area to be as small as possible-- but PCB routing wants to mess that up.
The larger the loop area, the more RF problems you will have. Not only will you emit more RF than you want, but you will also receive more RF.
The signals on the bottom (blue) layer will want their return path to be on the adjacent plane on the next layer (cyan)-- since that makes the loop area as small as possible. Signals on the top (red) layer will have their return path on the gold layer.
If a signal starts on the top layer then goes through a via to the bottom layer then the signal return path will want to switch from the gold to cyan layers, at the point of the via! This is a major function of decoupling caps. Normally one plane would be GND and the other would be VCC. A signal return path can go through the decoupling cap when switching between planes. That is why it is often important to have caps between planes even when it is not obviously needed for power reasons.
Without a decoupling cap between planes, the return path cannot take a more direct route and so the loop area increases in size-- and EMC problems increase.
But voids/splits in the planes can be even more problematic. Your gold layer has split planes, and signal traces, which create problems. If you compare the red and gold layers you will see how signals cross the voids in the planes. Every time a signal crosses a void in the plane then something is going to go bad. The return current is going to be on the plane, but it can't follow the trace across the void so it has to take a major detour. This increases loop area and your EMC problems.
You can place a cap across the void, right where the signals cross. But a better approach would be to reroute things to avoid this in the first place.
Another way the same problem can be created is when you have several vias that are close together. The clearance between the vias and the plane can create slots in the planes. Either decrease the clearance, or spread the vias out so a slot does not form.
Ok, so that's the biggest issue with your board. Once you understand that then you have to look at signal termination and controlling trace impedance. After that, you have to look at shielding and chassis GND issues with your Ethernet connection (not enough info in the Q to comment accurately).
I hope that helps. I really breezed by the issues but that should get you going.
Best Answer
The solenoid doesn't care if you supply it PWM or a clean analogue voltage so you could use an LC low pass filter to "convert" PWM to a smoother DC value. Values depend on dynamic expectations of your solenoid and PWM frequency. Fit it at the driving end of the cable to reduce EM emissions on the cable.
Note that using a twisted pair won't eliminate e-fields being generated by the cable because it is not a balanced output drive due to one wire being 0V. Add a screen or use coax for better results.