In a reaction carried out at 500 K, 0.001% of the total number of collisions are effective. The energy of activation of the reaction is approximately 

The fraction  of molecules having energy equal to or greater than $E_a$ is:

$x=\frac{n}{N}=e^{-E_a/RT}(x=\frac{n}{N}=0.001\%=\frac{0.001}{100}={10}^{-5})$

$\therefore \log x=\frac{-E_a}{2.3\times RT}(R=2calmo{l}^{-1}{K}^{-1},T=500K)$

$\log {10}^{-5}=\frac{-E_a}{2.3\times 2\times 500}$

$E_a=11.5\times 103calmo{l}^{-1}=11.5kcalmo{l}^{-1}$