Not quite - your reasoning only goes half way. Yes, the probability of there being a black in the bag and you pulling a white is 1/4. But there is also a chance that there was white on the bag (1/2) and you pull white (1), for a total of 1/2. Hence 3/4 of the time you will pull a white marble: 1 out of 4 times when there was a black in the bag and 2 out of 4 times when there is a white. Hence having pulled a white, the probability that there was a white to start is (2/4)/(2/4+1/4) = 2/3.

Alternatively you can use Bayes' Theorem. If we let P(W_1) = probability that the original ball in the bag is white, and P(W_2) = probability that the ball drawn from the bag is white, then the probability that the original ball is white given that you drew a white from the bag is:

$ P(W_1|W_2) = \frac {P(W_2|W_1)P(W_1)}{P(W_2)} = \frac {(1)(1/2)}{(3/4)} = \frac 2 3 $