The time required for completion of 99.9% of a 1^st order reaction is 25 min.  Its ${\mathrm{t}}_{\frac{1}{2}}$ is

$$\begin{gathered} \frac{t_{99.9\%}}{t_{50\%}}=\frac{{\displaystyle \frac{2.303}{k}}\log {\displaystyle \frac{a}{a-0.999a}}}{\frac{2.303}{k}\log \frac{a}{a-0.5a}} \\ \frac{t_{99.9\%}}{t_{50\%}}=\frac{{\displaystyle \log \frac{1}{0.001}}}{\log {\displaystyle \frac{1}{0.5}}} \\ {t}_{99.9\%}=10\times t_{50\%} \\ {t}_{50\%}=\frac{25}{10}=2.5 \end{gathered}$$