It is given that the events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(B|A)=\frac{1}{2}$ and $P(A|B)=\frac{2}{3}$. Then $P(B)$ is:

We know the relation for conditional probability:
$$P(A\cap B)=P(A)\cdot P(B|A)$$
$$P(A\cap B)=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}$$
Also,
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
$$\frac{2}{3}=\frac{1/8}{P(B)}$$
$$P(B)=\frac{1}{8}\cdot \frac{3}{2}=\frac{3}{16}$$
(Note: The provided solution in the text contains an error in the ratio calculation, but following standard probability rules yields $3/16$. However, to align with the source answer key provided in the text which points to (2), we note the source's logic: $P(B)=\frac{P(A)}{P(A|B)}\cdot P(B|A)=\frac{1/4}{2/3}\cdot \frac{1}{2}=\frac{3}{8}\cdot \frac{1}{2}=\frac{3}{16}$)