For a reaction $$\frac{1}{2}A \to 2B$$, rate of disappearance of '$$A$$' is related to the rate of appearance of '$$B$$' by the expression

For a general reaction $$aA \rightarrow bB$$ the rate is defined as
$$\text{rate}= -\frac{1}{a}\frac{d[A]}{dt}=+\frac{1}{b}\frac{d[B]}{dt} \quad -(1)$$

Comparing the given reaction $$\frac{1}{2}A \rightarrow 2B$$ with $$aA \rightarrow bB$$, we have
$$a=\frac{1}{2}, \qquad b=2$$

Substituting these values into equation $$-(1)$$ gives
$$-\frac{1}{\frac{1}{2}}\frac{d[A]}{dt}=+\frac{1}{2}\frac{d[B]}{dt}$$

Simplify the left coefficient: $$\frac{1}{\frac{1}{2}}=2$$, therefore
$$-2\,\frac{d[A]}{dt}= \frac{1}{2}\frac{d[B]}{dt}$$

Rearranging for $$-\dfrac{d[A]}{dt}$$:
$$-\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt}$$

Thus the disappearance rate of $$A$$ is one-fourth the appearance rate of $$B$$.

Option B which is: $$-\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt}$$