Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6 the game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if he doesn't roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

It isn't that difficult, but many get it wrong, yet are absolutely convinced they're right. Don't be like them if you can help it. Note that it's legal to be bombed by the club with my answer, but not the other way around.

The question asked in a way to eliminate interpretation: What is the implied probability of bob winning on his second throw, given that he won the game? Or P(bob wins on his second throw|bob wins).

Because, you know, the way the question is asked originally, the answer is really 0 or 1, depending on what happened.
Anyway, both your answers are wrong. The first one doesn't even make sense. What would the meaning of bob and sue be in the value?

I just guessed and of course it was completely wrong:
i thought when Bob has only 2 turns and there are 6 possibilities because of the die his chance is still 1/6

What we want to know is the probability Bob ended the game on his second turn ("Bob2"), given that he ended it ("Bob"). That probability is probability(intersection(Bob2, Bob))/probability(Bob) (AKA P(Bob2|Bob) like Z-Man said). Bob2 is Bob ending the game on his second turn; Bob is Bob ending it on any of turn. Bob2 implies Bob per definition, so this simplifies to probability(Bob2)/probability(Bob).

Moving along, we see that the probability of ending on any turn is 1/6, and that this scales the probability of all subsequent turns by 5/6. This makes the probability of a game having ended on turn n (1/6)*(5/6)^n, where n=0 is the first. Probability(Bob2) is then easily shown to be 125/1296. Probability(Bob) is the sum of all (1/6)*(5/6)^n for positive odd n, which is 5/11 (somebody else show that). Now plug them in and you get (125/1296)/(5/11)=275/1296.