detailed solution

Correct option is B

According Arrhenius,

$\text{Rate constant, }k{\text{ = Ae}}^{-\text{Ea}/\text{RT}}$

$k_{\text{1}}={\text{Ae}}^{-{\text{Ea}}_{\text{1}}\text{/RT}}$

$k_{\text{2}}={\text{Ae}}^{-{\text{Ea}}_{\text{2}}\text{/RT}}$

$\frac{k_1}{k_2}=\frac{\cancel{\text{A}}{\text{e}}^{-E{a}_1/RT}}{\cancel{\text{A}}{\text{e}}^{-E{a}_2/RT}}$

$\Rightarrow \frac{2}{1}=\frac{e^{-E{a}_1/RT}}{e^{-E{a}_2/RT}}$

$e^{-E{a}_1/RT}=2e^{-E{a}_2/RT}$

Apply ln on both sides then we will get

$\ln {\text{e}}^{-\frac{{\text{E}}_{{\text{a}}_1}}{\text{RT}}}=\ln 2{\text{e}}^{-\frac{{\text{E}}_{{\text{a}}_2}}{\text{RT}}}$

$\ln {\text{e}}^{-\frac{{\text{E}}_{{\text{a}}_1}}{\text{RT}}}=\ln 2+\ln {\text{e}}^{\left(-\frac{{\text{E}}_{{\text{a}}_2}}{\text{RT}}\right)}$         $\left(\log \text{ab}=\log \text{a}+\log \text{b}\right)$

$\ln {\text{e}}^{\frac{-{\text{E}}_{\text{a}}{}_{\text{1}}}{\text{RT}}}-\ln {\text{e}}^{\frac{-{\text{E}}_{{\text{a}}_{\text{2}}}}{\text{RT}}}=\ln 2$          

${\ln }_{\text{e}}\frac{{\text{e}}^{\frac{-{\text{E}}_{{\text{a}}_{\text{1}}}}{\text{RT}}}}{{\text{e}}^{\frac{-{\text{E}}_{{\text{a}}_{\text{2}}}}{\text{RT}}}}=\ln 2$               $\left(\log \text{a}-\log \text{b}=\log \frac{\text{a}}{\text{b}}\right)$         

${\ln }_{\text{e}}{\text{e}}^{\frac{-{\text{E}}_{{\text{a}}_{\text{1}}}}{\text{RT}}+\frac{{\text{E}}_{{\text{a}}_{\text{2}}}}{\text{RT}}}={\ln }_{\text{e}}2$            $\left(\log \frac{a^m}{a^n}=\log a^{m+n}\right)$

Remove ln on both sides

$\frac{-{\text{E}}_{\text{a1}}}{\text{RT}}+\frac{{\text{E}}_{{\text{a}}_{\text{2}}}}{\text{RT}}=2$                  

$-{\text{E}}_{{\text{a}}_{\text{1}}}+{\text{E}}_{{\text{a}}_{\text{2}}}=\underset{\begin{array}{c}\downarrow \\ +\text{ve}\end{array}}{2}\times \underset{\begin{array}{c}\downarrow \\ +\text{ve}\end{array}}{\text{R}}\underset{\begin{array}{c}\downarrow \\ +\text{ve}\end{array}}{\text{T}}$

${\text{E}}_{{\text{a}}_{\text{2}}}-{\text{E}}_{{\text{a}}_{\text{1}}}=+\text{ve}$

${\text{E}}_{{\text{a}}_{\text{2}}}>{\text{E}}_{{\text{a}}_{\text{1}}}$

Hence  $E_1<E_2$