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

We interpret $$P\!\left(\frac{A}{B}\right)$$ and $$P\!\left(\frac{B}{A}\right)$$ as conditional probabilities $$P(A\mid B)$$ and $$P(B\mid A)$$ respectively.

Given data:
$$P(A)=\frac14,\qquad P(A\mid B)=\frac12,\qquad P(B\mid A)=\frac23$$

Using the definition of conditional probability
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\qquad -(1)$$
$$P(B\mid A)=\frac{P(A\cap B)}{P(A)}\qquad -(2)$$

From equation $$-(1)$$:
$$P(A\cap B)=P(A\mid B)\,P(B)=\frac12\,P(B)\qquad -(3)$$

From equation $$-(2)$$ and the given $$P(A)=\frac14$$:
$$P(A\cap B)=P(B\mid A)\,P(A)=\frac23\cdot\frac14=\frac16\qquad -(4)$$

Equate the two expressions $$-(3)$$ and $$-(4)$$ for $$P(A\cap B)$$:
$$\frac12\,P(B)=\frac16$$

Solve for $$P(B)$$:
$$P(B)=\frac16\div\frac12=\frac13$$

Therefore $$P(B)=\frac13$$.

Option B which is: $$\frac{1}{3}$$