The rate of a reaction A doubles on increasing the temperature from...

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

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

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

$\dfrac{1}{300} + \dfrac{1}{310} = \dfrac{2}{T}$

$⇒ T = \dfrac{300 \times 310}{610} \times 2$

$= 304.92$

Hence, the temperature of reaction B should be increased from $300K$ by $304 . 92 - 300 = 4 . 92K$.