The rate of a reaction A doubles on increasing the temperature from   300   to   310   K…

$\log 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left\{\frac{1}{300}-\frac{1}{310}\right\}$ .......(i)

$\log 2=2\left[\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\mathrm{ }\left\{\frac{1}{300}-\frac{1}{\mathrm{T}}\right\}\right]$ ......(ii)

$\frac{2{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\mathrm{ }\left\{\frac{1}{300}-\frac{1}{\mathrm{T}}\right\}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\mathrm{ }\left\{\frac{1}{300}-\frac{1}{310}\right\}$

$\frac{1}{300}+\frac{1}{310}=\frac{2}{\mathrm{T}}$

$\Rightarrow \mathrm{ }\mathrm{T}=\frac{300\times 310}{610}\times 2$

$=304.92$

Hence, the temperature of reaction B should be increased from $300 \mathrm{K}$ by $304.92-300=4.92 \mathrm{K}$.