$\begin{aligned} & \log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.3 \times 8.3}\left(\frac{1}{200}-\frac{1}{300}\right) \\\\ & \log \frac{0.05}{0.03}=\frac{E_a}{2.3 \times 8.3}\left(\frac{1}{600}\right) \\\\ & \begin{aligned} (0.70-0.48)=\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600} \end{aligned} \\\\ & \begin{aligned} & \Rightarrow 0.22 =\frac{E_a}{2.3 \times 8.3} \times \frac{1}{600} \\\\ & \Rightarrow {E_a} = 2.3 \times 8.3 \times 600 \times 0.22 \\\\ & =2519.88 \\\\ & \approx 2520 \mathrm{~J} \end{aligned} \end{aligned}$