For a first order reaction:
$\Rightarrow [\mathrm{A}{]}_{\mathrm{t}}=[\mathrm{A}{]}_0{\mathrm{e}}^{-\mathrm{kt}}$

For $\mathrm{A}$ : Let $[\mathrm{A}{]}_{\mathrm{t}}$ be $\mathrm{y}$ and $[\mathrm{A}{]}_0$ be x
For a first order reaction:
 $\therefore \mathrm{k}=\frac{\ln 2}{{\mathrm{t}}_{1/2}}=\frac{\ln 2}{15 \min }$

$\Rightarrow \mathrm{y}={\mathrm{xe}}^{-\mathrm{kt}}$

$={\mathrm{xe}}^{-\left(\frac{\ln 2}{15}\right)\mathrm{t}}$    ....(1)

Similarly For B:
 $[\mathrm{B}{]}_{\mathrm{t}}=[\mathrm{B}{]}_0{\mathrm{e}}^{-\mathrm{kt}}$

Let $$\begin{gathered} [\mathrm{B}{]}_{\mathrm{t}}=\mathrm{y};[\mathrm{B}{]}_0=4\mathrm{x}; \\ \mathrm{k}=\frac{\ln 2}{{\mathrm{t}}_{1/2}}=\frac{\ln 2}{5 \min } \end{gathered}$$

$\Rightarrow \mathrm{y}=4{\mathrm{xe}}^{-\left(\frac{\ln 2}{5}\right)\mathrm{t}}$      ....(2)
Equating equation(1) and (2) we get:

$\Rightarrow {\mathrm{xe}}^{-\left(\frac{\ln 2}{15}\right)\mathrm{t}}=4{\mathrm{xe}}^{-\left(\frac{\ln 2}{5}\right)\mathrm{t}}$

$\Rightarrow \mathrm{t}\times \left[\frac{\ln 2}{5}-\frac{\ln 2}{15}\right]=\ln 4$

$\Rightarrow \mathrm{t}\times \ln 2\left[\frac{1}{5}-\frac{1}{15}\right]=2 \ln 2$

$\Rightarrow {\mathrm{e}}^{t\left(\frac{\ln 2}{5}-\frac{\ln 2}{15}\right)}=4$

$\Rightarrow \mathrm{t}=15 \min$