Activation energy of any chemical reaction can be calculated if one knows the value of

The correct answer is rate constant at two different temperatures.

To understand why, we can refer to the Arrhenius equation, which describes the quantitative relationship between the rate constant of a chemical reaction and the absolute temperature:
$k=A{e}^{-E_a/RT}$
where:
• $k$ is the rate constant of the reaction,
• $A$ is the pre-exponential factor (or frequency factor),
• $E_a$ is the activation energy,
• $R$ is the universal gas constant, and
• $T$ is the absolute temperature (in Kelvin).

Taking the natural logarithm of both sides of the Arrhenius equation:
$\ln k=\ln A-\frac{E_a}{RT}$
If we determine the rate constant at two different temperatures, say $T_1$ and $T_2$, the corresponding rate constants are $k_1$ and $k_2$ respectively. Writing the logarithmic form for both conditions:
$\ln k_1=\ln A-\frac{E_a}{R{T}_1}$
and
$\ln k_2=\ln A-\frac{E_a}{R{T}_2}$
By subtracting the equation for $k_1$ from the equation for $k_2$, we eliminate the unknown pre-exponential factor $A$:
$\ln k_2-\ln k_1=\left(\ln A-\frac{E_a}{R{T}_2}\right)-\left(\ln A-\frac{E_a}{R{T}_1}\right)$
Simplifying the relationship:
$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
Converting the natural logarithm (ln) to base-10 logarithm (log):
$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

From this final integrated expression, we can conclude that the activation energy ($E_a$) can be calculated if we know the rate constant at two different temperatures.