Explanation

(c) After $n$ half-lives the quantity of a radioactive substance left intact (undecayed) is given by
$\ \ \ \ \ \ \ N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\ \ \ \ \ \ = \ N_{0}\left( \frac{1}{2} \right)^{t/T_{1/2}}$
Here, $N = \ \frac{1}{16}N_{0},\ t = 2h$
$\frac{1}{16}N_{0} = N_{0}\left( \frac{1}{2} \right)^{\frac{2}{T_{\frac{1}{2}}}}\ $
$\ \ \ \ \ \ \ \ \ \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{2/T_{1/2}}$
Equating the powers on both sides
$\ \ \ \ \ \ 4\ = \frac{2}{T_{1/2}}$
$\ \ \ \ \ \ \ \ \ T_{1/2} = \frac{1}{2}h = \ 30$ min