Hint: We can model this by arranging the ten balls in a line, so that the first four are those moved into the second urn then removed in that order. Given that the first ball is white, what is the probability that the second ball is too?
That is the short method.
The long method is:
Let $E_x$ be the event of $x$ white balls among the four balls drawn from five white and five black, for $x\in\{0,1,2,3,4\}$. This corresponds to a hypergeometric distribution.
$$\mathsf P(E_x) ~=~ \dfrac{\dbinom 5 x\dbinom 5 {4-x}}{\dbinom {10}4}\cdot\mathbf 1_{x\in\{0,1,2,3,4\}}$$
The events $E_0,E_1,E_2,E_3,E_4$ are thus not equally probable.
Let $A, B$ be the event of drawing white balls in two consecutive draws from among those four.
$$\mathsf P(B\mid A)=\dfrac{\sum_{x=2}^4 \mathsf P(E_x)~\mathsf P(A\cap B\mid E_x)}{\sum_{x=1}^4 \mathsf P(E_x)~\mathsf P(A\mid E_x)}=\dfrac{\sum_{x=2}^4 \binom 5x\binom 5{4-x}\binom x 2/\binom 4 2}{\sum_{x=1}^4 \binom 5x\binom 5{4-x}\binom x 1/\binom 4 1}$$
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:$$\frac23\frac12+\frac13\frac13=\frac49$$
Best Answer
We are looking for the probability that the number of white balls is 2 given that we have already selected a white ball. So, we are looking for \begin{equation} \frac{\text{Pr}(X = 2 \cap B = w)}{\text{Pr}(B=w)} \end{equation} Where $X$ is the number of white balls and $B$ is the ball drawn. It is given to us that $\text{Pr}(X = i) = \lambda \cdot i^2$. We can see that \begin{equation*} \text{Pr}(X = 2 \cap B = w) = (\lambda \cdot 2^2) \cdot \frac{2}{n} = \frac{8\lambda}{n} \end{equation*}
Now, to find $\text{Pr}(B=w)$, we need the sum \begin{equation*} \sum_{i=1}^n (\lambda\cdot i^2) \cdot \frac{i}{n} \end{equation*} This is the sum of the probabilities that the number of white balls is $i$ and we choose a white ball. Finally, we substitute these into the expression above so that we have \begin{equation} \frac{\frac{8\lambda}{n}}{\sum_{i=1}^n (\lambda\cdot i^2) \cdot \frac{i}{n}} = \frac{8}{\sum_{i=1}^n i^3} = \frac{8}{\frac{1}{4}(n(n+1))^2} = \frac{32}{(n(n+1))^2} \end{equation} This is the probability that number of white balls is 2 given that we have chosen a white ball.