The probabilities that we have $(1,2,3,4)$ white balls in the bag are
$$\left({1\over8},{3\over8},{3\over8},{1\over8}\right)\ .$$
Given that we have $(1,2,3,4)$ white balls in the bag the probabilities that we draw two white balls are
$$\left(0,{1\over6}, {1\over2},1\right)\ .$$
The overall probability $p(2w)$ to draw two white balls then is given by
$$p(2w)={1\over8}\cdot0+{3\over8}\cdot{1\over6}+{3\over8}\cdot{1\over2}+{1\over8}\cdot1={3\over8}\ .$$
Finally the probability that all three balls are white computes to
$$p(3w)={1\over8}\cdot0+{3\over8}\cdot0+{3\over8}\cdot{1\over4}+{1\over8}\cdot1={7\over32}\ .$$
It follows that the conditional probability $p(3w|2w)$ is given by
$$p(3w|2w)={p(3w\wedge 2w)\over p(2w)}={p(3w)\over p(2w)}={7\over12}=0.58333\ldots\ .$$