1. Use Arrhenius equation

For a reaction, $k = A e^{-E_a/(RT)}$

So for two temperatures, the ratio of rate constants is $\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Since rate doubles, we take $\frac{k_2}{k_1}=2$

2. For reaction A

Given reaction A doubles when temperature increases from $300$ K to $310$ K.

Thus, $\ln 2 = \frac{E_A}{R}\left(\frac{1}{300}-\frac{1}{310}\right)$

3. For reaction B

Activation energy of B is twice that of A: $E_B = 2E_A$

Let the required temperature rise for B from $300$ K be $\Delta T$. Then final temperature is $300+\Delta T$

Again rate doubles, so $\ln 2 = \frac{E_B}{R}\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

Substitute $E_B=2E_A$: $\ln 2 = \frac{2E_A}{R}\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

4. Compare the two equations

From reaction A: $\ln 2 = \frac{E_A}{R}\left(\frac{1}{300}-\frac{1}{310}\right)$

From reaction B: $\ln 2 = \frac{2E_A}{R}\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

Equating, $\frac{E_A}{R}\left(\frac{1}{300}-\frac{1}{310}\right)=\frac{2E_A}{R}\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

Cancel $\frac{E_A}{R}$: $\left(\frac{1}{300}-\frac{1}{310}\right)=2\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

Now, $\frac{1}{300}-\frac{1}{310}=\frac{10}{300\cdot 310}=\frac{1}{9300}$

So, $\frac{1}{9300}=2\left(\frac{1}{300}-\frac{1}{300+\Delta T}\right)$

Hence, $\frac{1}{300}-\frac{1}{300+\Delta T}=\frac{1}{18600}$

5. Solve for $\Delta T$

$\frac{1}{300}-\frac{1}{300+\Delta T} = \frac{\Delta T}{300(300+\Delta T)} = \frac{1}{18600}$

So, $\frac{\Delta T}{300(300+\Delta T)}=\frac{1}{18600}$

Cross-multiply: $18600\Delta T = 300(300+\Delta T)$

$18600\Delta T = 90000 + 300\Delta T$

$18300\Delta T = 90000$

$\Delta T = \frac{90000}{18300} \approx 4.92\text{ K}$

6. Check options

The required increase in temperature is $4.92\text{ K}$

So the correct option is B.