To solve this problem, we need to use the Arrhenius equation and the concept of temperature dependence of reaction rates.$\newline$The Arrhenius equation is given by:$\newline k = A e^{-\frac{E_a}{RT}} \newline$where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy,$\newline R$ is the universal gas constant, and $T$ is the temperature in Kelvin.$\newline$The ratio of rate constants at two different temperatures can be expressed as:$\newline \frac{k_2}{k_1} = e^{-\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \newline$Given that the rate quadruples, we have:$\newline \frac{k_2}{k_1} = 4 \newline$Let's convert the temperatures from Celsius to Kelvin:$\newline T_1 = 27^\circ C = 300 \,$K$\newline T_2 = 57^\circ C = 330 \,$K$\newline$Substitute the values into the equation:$\newline 4 = e^{-\frac{E_a}{8.314} \left( \frac{1}{330} - \frac{1}{300} \right)} \newline$Take the natural logarithm on both sides:$\newline \ln(4) = -\frac{E_a}{8.314} \left( \frac{1}{330} - \frac{1}{300} \right) \newline$Given $\log 4 = 0.6021,$ convert to natural log:$\newline \ln(4) = 0.6021 \times 2.303 = 1.386 \newline$Substitute back:$\newline 1.386 = -\frac{E_a}{8.314} \left( \frac{1}{330} - \frac{1}{300} \right) \newline$Calculate the difference in reciprocals:$\newline \frac{1}{330} - \frac{1}{300} = \frac{300 - 330}{330 \times 300} = -\frac{30}{99000} = -\frac{1}{3300} \newline$Substitute this into the equation:$\newline 1.386 = \frac{E_a}{8.314} \times \frac{1}{3300} \newline$Solve for $E_a: \newline E_a = 1.386 \times 8.314 \times 3300 \newline E_a = 38044.6 \,$J/mol$\newline$Convert to kJ/mol:$\newline E_a = 38.0446 \,$kJ/mol$\newline$Therefore, the activation energy is approximately $38.04 \,$kJ/mol.$\newline$This corresponds to Option 3.