Relation between temperature and activation energy is given by Arrhenius equation. 

$\mathrm{k} ={\mathrm{Ae}}^{-\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\right)}$

$$\begin{gathered} \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1} ={\mathrm{e}}^{-\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}+\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}\right)} \\ \ln \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)=-\left(\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}+\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}\right) \\ \ln \left(2\right) =\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ \log \left(2\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{300}-\frac{1}{600}\right) \\ 0.3010=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{1}{600}\right) \\ {\mathrm{E}}_{\mathrm{a}}=3.45 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$