At $t=0,\left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}$
$\frac{N_{A}}{N_{B}}=\left(\frac{1}{e}\right)^{2}$
According to radioactive decay, $\frac{N}{N_{0}}=e^{-\lambda t}$
$\therefore \,\frac{N_{A}}{\left(N_{0}\right)_{A}} =e^{-\lambda_{A} t} $ ..... $(i)$
$\frac{N_{B}}{\left(N_{0}\right)_{B}} =e^{-\lambda_{B} t}$ ..... $(ii)$
Divide $(i)$ by $(ii)$, we get
$\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}\right) t}$ or, $\frac{N_{A}}{N_{B}}=e^{-(5 \lambda-\lambda) t}$
or, $\left(\frac{1}{e}\right)^{2}=e^{-4 \lambda t}$ or, $\left(\frac{1}{e}\right)^{2}=\left(\frac{1}{e}\right)^{4 \lambda t}$
or, $4 \lambda t=2$ $ \Rightarrow \quad t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}$
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