The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:
(Assume Activation energy and pre-exponential factor are independent of temperature;
ln(2) = 0.693; R = 8.314 J mol$$^{-1}$$ K$$^{-1}$$)

We start with the Arrhenius equation, which relates the rate constant $$k$$ to the activation energy $$E_a$$:

$$k = A\,e^{-E_a/(RT)}$$

Here $$A$$ is the pre-exponential factor, $$R$$ is the gas constant and $$T$$ is the absolute temperature. Because $$A$$ and $$E_a$$ are assumed to be independent of temperature, we can take the natural logarithm of the ratio of two rate constants at two different temperatures to eliminate $$A$$. Thus, for temperatures $$T_1$$ and $$T_2$$:

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

For the given reaction the temperature rises from $$T_1 = 300\ \text{K}$$ to $$T_2 = 310\ \text{K}$$ and the rate quadruples, so $$k_2/k_1 = 4$$. Substituting these values we get:

$$\ln(4)=\dfrac{E_a}{R}\left(\dfrac{1}{300}-\dfrac{1}{310}\right)$$

We first evaluate the left-hand side. Since $$4 = 2^2$$,

$$\ln(4)=2\ln(2)=2(0.693)=1.386$$

Next we calculate the temperature difference term:

$$\dfrac{1}{300}-\dfrac{1}{310}=\dfrac{310-300}{300\times310}=\dfrac{10}{93000}=\dfrac{1}{9300}\approx 1.07527\times10^{-4}\ \text{K}^{-1}$$

Now we substitute $$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$$ and rearrange the equation to solve for $$E_a$$:

$$E_a = \dfrac{R\ln(4)}{\dfrac{1}{300}-\dfrac{1}{310}}$$

$$E_a = \dfrac{8.314 \times 1.386}{1.07527\times10^{-4}}$$

First compute the numerator:

$$8.314 \times 1.386 = 11.526$$

Now divide by the denominator:

$$E_a = \dfrac{11.526}{1.07527\times10^{-4}}\ \text{J mol}^{-1}$$

$$E_a \approx 1.07206\times10^{5}\ \text{J mol}^{-1}$$

To express this in kilojoules per mole, we divide by 1000:

$$E_a \approx 107.2\ \text{kJ mol}^{-1}$$

Hence, the correct answer is Option C.