Answer
$P(Y_{2}|X)=\frac{.3\times .4}{.2\times .3+.3\times .4 + .6\times .3}\approx .33$
Work Step by Step
According to Bayes' theorem:
$P(A_{1}|B)=\frac{P(B|A_{1})P(A_{1})}{P(B|A_{1})P(A_{1})+P(B|A_{2})P(A_{2})+P(B|A_{3})P(A_{3})}$
Where $A_{1}, A_{2}, A_{3}$ form a partition of $S$.
Meaning that $P(A_{1})+P(A_{2})+P(A_{3})=1$
Here, we have
$P(X|Y_{1})= .2$
$P(X|Y_{2})= .3$
$P(X|Y_{3})= .6$
$P(Y_{1})= .3$
$P(Y_{2})= .4$
Be aware, that X and Y equal to B and A in the definition of the theorem, respectively.
We know, that $P(Y_{3})=1-P(Y_{1})-P(Y_{2})= 1-.3-.4=.3$
We can substitute into the definition:
$P(Y_{2}|X)=\frac{.3\times .4}{.2\times .3+.3\times .4 + .6\times .3}\approx .33$
