For a first-order decomposition reaction, the half-life period...

For a first-order reaction, the rate constant is inversely proportional to the half-life: $k = \dfrac{0.693}{t_{1/2}}$.

At $T_1 = 27^{\circ}\mathrm{C} = 300 \text{ K}$, $t_{1/2} = 40 \text{ min}$, so $k_1 = \dfrac{0.693}{40}$.

At $T_2 = 37^{\circ}\mathrm{C} = 310 \text{ K}$, $t_{1/2} = 20 \text{ min}$, so $k_2 = \dfrac{0.693}{20}$.

Taking the ratio of the rate constants:

$\dfrac{k_2}{k_1} = \dfrac{40}{20} = 2$

Using the Arrhenius equation:

$\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.303 R} \left( \dfrac{T_2 - T_1}{T_1 T_2} \right)$

Substituting the given values:

$\log 2 = \dfrac{E_a}{2.303 \times 8.314} \left( \dfrac{310 - 300}{300 \times 310} \right)$

$0.301 = \dfrac{E_a}{19.147} \left( \dfrac{10}{93000} \right)$

$E_a = \dfrac{0.301 \times 19.147 \times 93000}{10}$

$E_a = 53598 \text{ J mol}^{-1} = 53.6 \text{ kJ mol}^{-1}$

Answer: $53.6 \text{ kJ mol}^{-1}$