The relation between rate constant at two different temperatures and activation energies is given by :

$\log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$

So,

 $$\begin{gathered} \log \left(\frac{2.60\times {10}^{-3}}{2.60\times {10}^{-5}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left[\frac{1}{400}-\frac{1}{500}\right] \\ {\mathrm{E}}_{\mathrm{a}}= \log \left({10}^2\right)\times 2.303\times 8.314\left[\frac{400\times 500}{100}\right] \\ {\mathrm{E}}_{\mathrm{a}}=2\times 2.303\times 8.314\times \frac{200000}{100} \\ {\mathrm{E}}_{\mathrm{a}}=76.6 \mathrm{KJ}/\mathrm{mol} \end{gathered}$$