According to the Arrhenius equation, if $k_1$ and $k_2$ are the rate constants at absolute temperatures $T_1$ and $T_2$ ($T_2 > T_1$), then $\log\dfrac{k_2}{k_1}$ equals:

Taking $\log_{10}$ of the Arrhenius expression at the two temperatures and subtracting gives $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right) = \dfrac{E_a}{2.303\,R}\cdot\dfrac{T_2 - T_1}{T_1 T_2}$. Option A is missing the $2.303$ factor needed for $\log_{10}$; option C has the wrong sign for $T_2 > T_1$.