What do you mean by the correct answer, though? I mean, in a "prove that" question the correct answer includes the full working to derive the final result, not just the final result itself.
It's basically only two (or possibly three, depending on how small your writing is) lines of working:
n sweets in a bag, 6 orange. Hence probability of drawing one orange is 6/n, and probability of drawing a second orange sweet after the first is (6-1)/(n-1) = 5/(n-1). Overall probability is therefore:
(6/n)*(5/(n-1)) = (6*5)/(n(n-1)) = 30/(n^2 -n).
Told that this equals a third, ie: 30/(n^2 -n) = 1/3, or, rearranging 90 = n^2-n, or n^2-n-90=0.
Not a difficult problem in principle, requiring simply some level of competence over algebra and the ability to apply the basic rules of probability with one unknown quantity -- and then to put it all together, which is to be fair a highly non-trivial skill. But then that's why this is exactly the sort of problem that should be on an exam. It separates those who actually know the maths from those who merely are able to regurgitate results without really knowing what they all mean.