$0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60}$    $...(2)$

$(1) /(2)$

$\frac{4}{3}=\mathrm{e}^{600 \mathrm{k}(2-1)}$

$\frac{4}{3}=\mathrm{e}^{600 \mathrm{k}}$

$\ln \frac{4}{3}=600 \mathrm{k}$

$\ln \frac{4}{3}=600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}}$

$\mathrm{t}_{1 / 2}=600 \frac{\ln 2}{\ln \frac{4}{3}}$sec

$\mathrm{t}_{1 / 2}=600 \times \frac{\log 2}{\log 4-\log 3} \text { sec. }=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min}$

$\mathrm{t}_{1 / 2}$

Ans. $24$