For this question you simply have to use bayes' rule:
$$ P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)} $$
First let us see what is the probability of selecting any of the urns. Since the urns are indistinguishable and there are 4 of them, the probabilities of selecting any of the urns is:
$$P(U_1) = P(U_2) = P(U_3) = P(U_4) = \frac{1}{4}$$
where $U_i$ represents the $i^{th}$ urn.
We know one urn has all black balls, a second has 2 black and 1 white, a third has 2 white and 1 black, and the final one has 3 white ones. Counting how many white balls and black balls there are and looking at their proportion relative to the total number of balls, we find that the probability of selecting a white or black ball is:
$$P(ball_{white}) = P(ball_{black}) = \frac{3 +2 + 1}{4 \cdot 3} = \frac{6}{12} = \frac{1}{2}$$
Your problem asks this: "Given that the ball selected was white, what is the probability that it came from the urn containing only white balls". Without loss of generality, let us assume that $U_4$ is the urn that contains white balls only. You want to find:
$$P(U_4 \mid ball_{white})$$
Now we can use bayes' rule to rewrite the problem into one that is more suitable as follows:
$$P(U_4 \mid ball_{white}) = \frac{P(ball_{white} \mid U_4) \, P(U_4)}{P(ball_{white})} $$
$P(ball_{white} \mid U_4) = 1$, because if you select a ball from an urn that only contains white balls, you are guaranteed to get a white ball hence the probability is 1.
And we have already computed the probabilities $P(U_4)$ and $P(ball_{white})$ above. Substituting these values into the bayes' rule form of our problem, we get:
$$P(U_4 \mid ball_{white}) = \frac{1 \cdot \frac{1}{4}}{\frac{1}{2}} = \frac{1}{4}\cdot \frac{2}{1} = \frac{1}{2} $$
Therefore, the answer to your problem is $\frac{1}{2}$.