Your list of $3$ possibilities is correct. But there is no reason to assert that these possibilities are equally likely.
The problem is poorly posed, since we are not given any information about how the balls were put into the urn. (This is technically called the prior distribution.) From the answers supplied, it is very likely that you are expected to assume that $0$ white, $1$ white, and so on up to $4$ white are (before we chose the $2$ balls) equally likely. But that is an assumption, and not even a particularly reasonable assumption.
To solve the problem carefully, we need to set up some machinery. Let $W$ be the event they are all white, and $F$ be the event the first two balls removed are white. We want $\Pr(W|F)$.
By using the defining formula $\Pr(W|F)=\dfrac{\Pr(W\cap F)}{\Pr(F)}$ for conditional probabilities, we can calculate a formal answer. If you wish, I can later complete that approach.
But let's do it more informally. We put balls into an urn $600$ times, where $120$ times we put in all white, $120$ times we put in $1$ white and $3$ black, and so on.
How many times roughly did we draw $2$ white? If we were drawing from a $2$ white $2$ black, the probability of $2$ white is $\frac{1}{2}\cdot\frac{1}{3}$, so we got $2$ white about $120/6=20$ times. If we are drawing from a $3$ white $1$ black, the probability of $2$ white is $\frac{3}{4}\cdot\frac{2}{3}$, so we got $2$ white about $60$ times. Finally, if we drew from an all white, we got $2$ white for sure, so $120$ times.
The total number of times we got $2$ white was $200$. Of these times, $120$ came from an all white situation. So the probability is $\frac{120}{200}=\frac{3}{5}$.
