PCR amplification and error propagation

Your teacher is indeed correct.

In the first round you would get two identical molecules of the dsDNA.

In the second round you would get 3 identical molecules and one molecular with an A substituted for a G in one of the strands. ie.

No error (3 of the 4 molecules):

------G-------
------C-------

One mismatch (1 of the 4 molecules):

------A-------
------C-------

So there are a total of 8 strands of DNA after the second round and one of those strands has the mismatch. 1 / 8 = 0.125 = 12.5%

In round 3 you would have 8 dsDNA molecules and only one of those 8 dsDNA molecules would have the mismatch.