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.
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.
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:$$\frac23\frac12+\frac13\frac13=\frac49$$