Milk turns sour at $40°C$ three times faster than at $0° C$. Hence, ${\mathbf{E}}_a$ in calories of process of turning of milk sour is:

Milk becomes sour three times as quickly at $40°C$ as at $0° C$.

Then $\frac{\text{ rate at }{40}^{\circ }C}{\text{ rate at }{0}^{\circ }C}=\frac{k_{{40}^{\circ }\mathrm{C}}}{k_{0^{\circ }\mathrm{C}}}$

$E_a=\frac{2.303R{T}_1{T}_2}{\left(T_2-T_1\right)}\log \left(\frac{k_{{40}^{\circ }C}}{k_{0^{\circ }C}}\right)$

$E_a=\frac{2.303\times 2\times 313\times 273\log \left(3\right)}{40}$

Hence, option A is correct.