Given,

For first order reaction,
(i) half-life $\left(t_{1 / 2}\right)$ at temperature $300 \mathrm{~K}=50 \mathrm{~s}$
(ii) half-life $\left(t_{1 / 2}\right)$ at temperature $400 \mathrm{~K}=10 \mathrm{~s}$
$$
\begin{aligned}
& \therefore \quad K_1(\text { at } 300 \mathrm{~K})=\frac{0.693}{50}=0.014 \mathrm{~s}^{-1} \\
& \text { and } K_2(\text { at } 400 \mathrm{~K})=\frac{0.693}{10}=0.07 \mathrm{~s}^{-1} \\
& \left.\qquad \because K=0.693 / t_{(1 / 2)} \text { for lst order reaction }\right]
\end{aligned}
$$
where, $K_1$ and $K_2$ are rate constant at $300 \mathrm{~K}$ and $400 \mathrm{~K}$ respectively.
Also, according to Arrhenius theory
$$
\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 \cdot T_2}\right]
$$
where, $E_a=$ activation energy
$$
\begin{aligned}
T_1 & =\text { temperature }(300 \mathrm{~K}) \\
T_2 & =\text { temperature }(400 \mathrm{~K}) \\
R & =\operatorname{gas} \text { constant }\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1}\right)
\end{aligned}
$$

Thus,
$$
\begin{aligned}
& \log \frac{0.07}{0.014}=\frac{E_a}{2.303 \times 8.314}\left[\frac{400-300}{400 \times 300}\right] \\
& \text { or, } \quad \log 5=\frac{E_a}{2.303 \times 8.314}\left[\frac{100}{400 \times 300}\right] \\
& \text { or, } \quad 0.70=\frac{E_a}{19.15 \times 1200} \\
& \text { or, } \quad E_a=0.7 \times 19.15 \times 1200 \\
& =16086 \mathrm{~J} \\
& =16.08 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& E_a \approx 16.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&
\end{aligned}
$$

Hence, option (c) is the correct answer.