Suppose that we do not return balls to the box. The plain probability that the $i$-th ball selected is white is $\frac{k}{N}$, exactly like in the case of returning the ball to the box.
One way of seeing this is to number the balls, white from $1$ to $k$, black from $k+1$ to $N$. Imagine that we draw the balls, one at a time, until all the balls are gone. All permutations of the labels are equally likely, and the fraction of these permutations for which the $i$-th ball drawn is white is $\dfrac{k}{N}$. (It can take a while until this fact becomes "obvious"!)
But in drawing one at a time, there are other probabilities that we can compute, for example the probability that the $3$rd ball drawn is white given that the first two were white. This conditional probability is not $\dfrac{k}{N}$, it is $\dfrac{k-2}{N-2}$ (except in trivial cases, like $k=1$).
Suppose again that we draw the balls, one at a time, until they are all gone. The conditional probability that the $3$rd ball drawn is white, given that the last two balls drawn (of the $N$) are white is also $\dfrac{k-2}{N-2}$. So it is not the temporal order of the drawing that matters in evaluating the conditional probability.
When we calculate a conditional probability, it is not the act of receiving information that matters, it is the act of using the information to calculate a conditional probability, that is, to restrict the sample space.
When one does replacement, conditional and unconditional probabilities are the same, since the sample space is unchanged.
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
Best Answer
Your probability is
$$\mathbb{P}(B|RR)=\frac{\frac{8}{10}\cdot\frac{7}{9}}{\frac{3}{10}\cdot\frac{2}{9}+\frac{8}{10}\cdot\frac{7}{9}}=\frac{56}{62}=\frac{28}{31}$$
as you expected to find
the chance of 50% selecting A or B is irrelevant being the same for the two boxes