Answer:$ 84\;kJmol^{-1}$
Taking natural logarithms of the Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$ we get:
$ln\; k=ln A-\large\frac{E_a}{RT}$ which we can compare to $y=mx+c$
$\Rightarrow m = -\large\frac{E_a}{R}$
$\Rightarrow E_a = -mR = - (-1.01 \times 10^{4}\;K) \times (8.314\; J/KĀ·mol) = 83971\;Jmol^{-1} = 84\;kJmol^{-1}$