The half-life period of a $1^{\text{st }}$ order reaction is 60 minutes. What percentage will be left over after 240 minutes ?

$$\begin{gathered} \text{For a first order reaction:}{\mathrm{t}}_{1/2}=\frac{\ln 2}{\mathrm{k}} \\ \begin{array}{rl} & \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}\\ & =\frac{\ln 2}{60}{\min }^{-1}\\ & \frac{\left[{\mathrm{A}}_{\mathrm{t}}\right]}{\left[{\mathrm{A}}_0\right]}\\ & ={\mathrm{e}}^{-\mathrm{kt}}\\ & ={\mathrm{e}}^{-4\ln 2}\\ & =2^{-4}\\ & =0.0625\end{array} \\ \text{Hence }6.25\mathrm{\%} \text{of reactant will be left after }240 \text{minutes} \end{gathered}$$

Therefore, the correct option is (A).