I preassume that all bags contain exactly $2$ balls and that at random a bag is chosen from which the $2$ balls are taken one by one (if wrong then please tell me, so that I can delete the answer).
The first bag contains $2$ white balls and the question can be rephrased as:"If a white ball was selected at first draw then what is the probability that this ball was located in the first bag?"
There are exactly $3$ white balls in total and each of them has equal probability to be the ball that was selected at first draw. $2$ of these balls are located in a bag that contains another white ball and $1$ of them is located in a bag that does not contain another white ball.
So the probability that one of the $2$ balls located in a bag that contains another white ball was selected by first draw equals $\frac23$.
This event is the same as the event that the second draw will result in a white ball.
edit1:
If the above interpretation is wrong and the second ball can be chosen out each of the $3$ bags then the probability that the second ball is white is $\frac25$.
This because at the second round there are $5$ balls in total (all having equal probability to be chosen) of which $2$ are white.
edit2
If both interpretations above are wrong and by the second round each bag has the same probability to be chosen then the following calculation:
The probability that after drawing the first ball (which appeared to be white) we are in situation $|W\mid WB\mid BB|$ (i.e. one bag contains a white ball, one contains a white and a black ball and the third contains $2$ black balls) is $\frac23$ (i.e. the probability that the first ball was taken from the bag containing $2$ white balls; see first interpretation for that).
In this situation the probability that the second balls is white is $\frac13\cdot1+\frac13\frac12+\frac13\cdot0=\frac12$.
The probability that after drawing the first ball we are in situation $|WW\mid B\mid BB|$ is $1-\frac23=\frac13$.
In this situation the probability that the second balls is white is $\frac13\cdot1+\frac13\cdot0+\frac13\cdot0=\frac13$.
We conclude that the probability that the second ball is white is:$$\frac23\frac12+\frac13\frac13=\frac49$$
We want to find the probability that the two black balls were extracted from bag 1 given that two black balls were extracted from a randomly chosen bag. Let $E$ be the event that two black balls were extracted and $F$ be the event that the balls were extracted from bag 1.
\begin{align*}
\Pr(F \mid E) & = \frac{\Pr(E \cap F)}{\Pr(E \cap F) + \Pr(E \cap F')}\\
& = \frac{\Pr(F)\Pr(E \mid F)}{\Pr(F)\Pr(E \mid F) + \Pr(F')\Pr(E \mid F')}\\
& = \frac{\frac{1}{2} \cdot \frac{\binom{3}{2}}{\binom{8}{2}}}{\frac{1}{2} \cdot \frac{\binom{3}{2}}{\binom{8}{2}} + \frac{1}{2} \cdot \frac{\binom{6}{2}}{\binom{10}{2}}}
\end{align*}
where the calculations are based on the assumption that the balls are selected without replacement.
If selection with replacement was intended, the author should have explicitly stated that. Under that interpretation, your answer would have been correct.
Best Answer
Let $S_i$ be the event that you selected bag $i$; $W,B$ that you selected a white or black ball. I assume that be 'randomly selected' they selecting either bag is equally likely. Then you are being asked \begin{align*} P(S_A|B)&= \frac{P(S_A,B)}{P(B)}\tag 1\\ &=\frac{P(B|S_A)P(S_A)}{P(S_A,B)+P(S_B,B)}\tag 2\\ &=\frac{P(B|S_A)P(S_A)}{P(B|S_A)P(S_A)+P(B|S_B)P(S_B)}\\ &=\frac{(3/5)(1/2)}{(3/5)(1/2)+(7/12)(1/2)}\\ &=\frac{36}{71}. \end{align*} where in $(1)$ I used Bayes' rule and in $(2)$ I used the product rule on the numerator, and the law of total probability on the denominator.