The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:
$$(R = 8.314$$ JK$$^{-1}$$ mol$$^{-1}$$ and log 2 = 0.301$$)$$

We start with the Arrhenius equation for the temperature dependence of a rate constant:

$$k = A\,e^{-\dfrac{E_a}{RT}}$$

where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

For two different temperatures $$T_1$$ and $$T_2$$ with their corresponding rate constants $$k_1$$ and $$k_2$$, we write the Arrhenius equation twice and then divide:

$$\dfrac{k_2}{k_1}= \dfrac{A\,e^{-\dfrac{E_a}{RT_2}}}{A\,e^{-\dfrac{E_a}{RT_1}}}=e^{-\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)}$$

Taking natural logarithm on both sides gives

$$\ln\!\left(\dfrac{k_2}{k_1}\right)=\dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

We are told that the rate doubles when the temperature rises from $$300\,\text{K}$$ to $$310\,\text{K}$$, so

$$\dfrac{k_2}{k_1}=2,\qquad T_1=300\,\text{K},\qquad T_2=310\,\text{K}$$

Substituting these values, we have

$$\ln 2=\dfrac{E_a}{R}\left(\dfrac{1}{300}-\dfrac{1}{310}\right)$$

First we evaluate the temperature difference term:

$$\dfrac{1}{300}-\dfrac{1}{310}= \dfrac{310-300}{300\times310}= \dfrac{10}{93000}= \dfrac{1}{9300}\,\text{K}^{-1}$$

Now we use $$R = 8.314\;\text{J K}^{-1}\text{mol}^{-1}$$ and convert $$\ln 2$$ from the given common logarithm. The problem provides $$\log_{10}2 = 0.301$$, and we know the relation $$\ln x = 2.303\log_{10}x$$. Hence

$$\ln 2 = 2.303 \times 0.301 \approx 0.693$$

Putting everything into the equation:

$$0.693 = \dfrac{E_a}{8.314}\left(\dfrac{1}{9300}\right)$$

Rearranging for $$E_a$$ gives

$$E_a = 0.693 \times 8.314 \times 9300$$

First multiply $$0.693$$ by $$8.314$$:

$$0.693 \times 8.314 \approx 5.762$$

Then multiply by $$9300$$:

$$E_a = 5.762 \times 9300 \,\text{J mol}^{-1} \approx 5.3605 \times 10^{4}\,\text{J mol}^{-1}$$

Finally convert joules to kilojoules (1 kJ = 1000 J):

$$E_a \approx \dfrac{5.3605 \times 10^{4}\,\text{J}}{1000} = 53.6 \,\text{kJ mol}^{-1}$$

Hence, the correct answer is Option C.