Let $A$ denote the event that you selected two white marbles.
Let $B$ denote the event that you selected a white-1 marble (from at least one of the bags).
The question asks, what is the probability that both marbles are white ($A$) given that at least one of the marbles selected was a white marble labeled as white-1 ($B$).
As per our definition of conditional probability, $P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{|A\cap B|}{|B|}$
$|B|$ is how many ways you could have selected at least one of the marbles to be white-1. By inclusion exclusion principle, there are $300 + 300 - 1 = 599$ number of ways this could have happened. (Count how many ways if white1 was selected from the left bag and anything from right, add how many ways if white2 was selected from right bag and anything from left, subtract how many ways to select it from both since we counted that scenario twice already)
$|A\cap B|$ is how many ways you selected two whites given that at least one of them was labled as white-1. Again by inclusion exclusion principle you have $150 + 150 - 1 = 299$ number of ways. (white-1 from left and any white from right, white-1 from right and any white from left, minus white-1 from left and white-1 from right)
Thus, $P(A|B) = \frac{299}{599}$
Notice what happens in the case of only two marbles in each bag, for $|B|$ it would have been $2+2-1 = 3$, and for $|A\cap B|$ it would have been $1+1-1 = 1$ giving the answer of $\frac{1}{3}$ as in the earlier problem. The probability in fact happened to go up. In fact, if we were to have number of marbles in the bags get incredibly large, you'll see the probability getting closer and closer to $50\%$.
