We are given that:

When T_{1} = 27 + 273 = 300 K

Let k_{1} = k

When T_{2} = 37 + 273 = 310 K

k_{2} = 2 k

Substituting these values the equation:

Log (k_{2} / k_{1})

= E_{a} / 2.303 R {(T_{2} – T_{1}) / T_{1} T_{2}}

We will get:

Log (2 k / k) = E_{a} / 2.303 x 8.314 {(310 – 300) / 300 x 310}

Log 2 = E_{a} / 2.303 x 8.314 x (10 / 300 x 310)

E_{a} = 53598.6 J mol^{-1}

E_{a} = 53.6 kJ mol^{-1}

Hence, the energy of activation of the reaction is 53.6 kJ mol^{-1}