In order that "the first success occurs on the second trial", there must be failure on the first trial and a success on the second. The probability of failure on the first attempt is 1- p and the probability of success on the second trial is (1- p)p. So the probability of "first success on the second trial is
Solve p- p^3= 0.1275 or p^3- p- 0.1275. You don't say how you got p= 0.6. It certainly does not satisfy that. I get between 9.0 and 9.1.