Electronic – Can 300 ADC damage a battery wire (4AWG) rated for 60VDC, 150ADC if 300ADC @5VDC is passed for 2 seconds

Battery heating can be roughly defined using its internal resistance. We can get a rough estimate of a battery's internal resistance using it's cold cranking amp rating. This is rated at -18C, which is close to -20C.

For a nominal 12V battery: $$ R = \frac{12V - 7.2V}{CCA} $$

Let's assume this is constant vs. current draw.

Let's assume at time t=0, the battery is completely insulated from the world and has a heat capacity similar to pure water.

The amount of energy required to heat 17 kg of water from say even -20C to -18C is: $$ E = 4.2 \frac{J}{g K} \cdot 17 kg \cdot (20^oC - 18^oC) = 142.8 kJ $$

For a 9A draw, this would take:

$$ t = \frac{E}{(9A)^2 \cdot R} $$

Suppose you have a 200 CCA battery. Then we can find t = 73 seconds.

Problems with this analysis: Obviously the full weight of the battery is not just the liquid inside. Thus the time required to heat the battery 2 degrees C would be less. However, a larger problem is that you probably don't have a perfectly insulated battery. The 9A draw only produces ~2W, which could easily be lost from the battery (the cables are extremely good conductors away from the battery).

At temperatures much colder you're going to have to start worrying about fluids freezing, thus many things could just stop working. You're much better off following proper car winterization guide.

Most of the top results of a search for "AWG wire current" reference the Handbook of Electronic Tables and Formulas, each giving the same numbers, presumably copied from the print edition. If you use a different piece of wire from the battery to each R-LED-LED string, and if R=180 ohms and LED drops 3V each, that leaves resistor current I = (9-3-3V)/180ohm = 33.3mA per string. Ten of these strings in parallel (for 20 LEDs total) gives 333mA total from the battery.

The table offers a spec for 30 AWG of 142mA "Maximum amps for power transmission" which "uses the 700 circular mils per amp rule, which is very very conservative." Meanwhile Wikipedia gives a fusing current for 30 AWG copper wire as 10A for 30 seconds. A 9V battery will not be able to source anything remotely like 10A, so melting the wire is a non-issue. You could probably get away with 30AWG if it's all you have, or want a thin wire bundle for some reason. Anything thicker would remove doubt about current carrying capacity and 26 AWG with table spec of 361mA max would already be above your total expected current.

Whether a 9V battery could actually deliver 330mA for a reasonable length of time is another matter. There is wide variation in chemistry, construction, and thus capacity available. Wikipedia says a lithium model has typical capacity of 1200mAh, giving you perhaps 3 hours runtime. Lithium 9V will run $5-10 each at retail, which may be more expensive than you had in mind. Alkaline will cost less, $3 each at retail, but also have less than half the capacity and also less current capability. An Energizer or Duracell alkaline might only give 1 hour runtime, and a cheapo model with low current capability might be lucky to light it up at all. If you want longer runtime, you'll need a physically larger battery which can store more energy.