In the following reaction

$x A \rightarrow y B$

$\therefore \quad \frac{-\mathrm{d} \mathrm{A}}{\mathrm{xdt}}=\frac{1}{\mathrm{y}} \frac{\mathrm{dB}}{\mathrm{dt}}$

$\frac{-\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}} \times \frac{\mathrm{x}}{\mathrm{y}}$

$\log \left[\frac{-\mathrm{d} \mathrm{A}}{\mathrm{dt}}\right]=\log \left[\frac{\mathrm{dB}}{\mathrm{dt}}\right]+\log \left(\frac{\mathrm{x}}{\mathrm{y}}\right)$

Comparing this equation with the equation given in question. We get,

$\log \frac{x}{y}=0.3010$ or $\log \frac{x}{y}=\log 2$

$\therefore \quad \frac{x}{y}=2$

$\therefore$ The reaction is of type $2 \mathrm{~A} \rightarrow \mathrm{B}$.

Hence, option (3) is correct.