The total marked rating of a cord-and attachment plug-connected room air conditioner shall not exceed 80 percent of the rating of a branch circuit where no other appliances are also supplied.
There are two ways you could solve the problem. Jerry gave the "trial & error" way, and I used the direct way.
Jerry started with the ampacity of the smallest conductor (15 amps) and multiplied by 0.80 to get a number (12 amps) that the A/C could not exceed on that conductor. Then he compared this to the actual A/C of 12.6 amps, said 15 would be too small, and bumped the conductor up to 20 amps. To check that, you'd then take 20 * 0.80 and get 16 amps ... compare that to 12.6 amps actual, and say the 20 amp conductor is OK.
I approached this from the opposite direction, starting at the A/C and heading for the minimum allowable conductor size, which means I need to divide the A/C by 0.80 (instead of multiplying the conductor by 0.80). Given the A/C load of (1512VA/120v) = 12.6 amps, dividing that by 0.80 gets me directly to 15.75 amps as the minimum conductor ampacity needed for that A/C ... no further calculations or checks are needed.
Formula is: (A/C amps) = (0.80) x (conductor amps)
All I did was re-arrange the formula to move the 0.80 from the conductor side to the A/C side. When you do this the formula becomes:
(A/C amps)/(0.80) = (conductor amps)
Either method gets you the correct answer. The only difference is that the way I did it gets you there faster. Do it whichever way makes the most sense for you.