Assuming that there is an equal probability that the coins are white or black (which I believe the question implies), and that the color of each coin is independent of each other, then there's a $\frac{1}{4}$ chance that they are both white.
If you attempt to apply Bayes' law here, then you would have the same probability, because both events are independent of each other.
$ P(A|B) = \frac{P(B | A) P(A)}{P(B)} $ $ P(A|B) = \frac{P(B)P(A)}{P(B)} $ $ P(A|B) = P(A) $
But perhaps the question is asking this: Assume the probability of a coin being either white or black. Given that the first two coins are black, what is the probability that the next two are white?
I think that the first result is more likely what they are asking, since there's many ways to do the second one. Regardless, there are some methods with the second one detailed here: http://en.wikipedia.org/wiki/Checking_whether_a_coin_is_fair
If you follow that, you could say that there is a $0%$ chance there are two whites remaining, because clearly, there's a 100% chance of getting black.
However, you could also say there is a $\frac{3}{4}$ chance of the coin being black, because the expected value is equal to 3/4. Then there is a $\dfrac{1}{16}$ chance.