Chemical Kinetics

For a reaction, the graph of $\ln k$ (on $y$-axis) and $\frac{1}{T}$ (on $x$-axis) is a straight line with a slope $-2 \times 10^4 \mathrm{~K}$. The activation energy of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is ( $R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )