Chemistry and Chemical Reactivity (9th Edition)

$E_a=84.9\ kJ/mol$
From the Arrhenius equation: $ln(k_2/k_1)=-E_a/R(1/T_2-1/T_1)$ $ln(3)=-E_a/8.314\ J/mol.K(1/310\ K-1/300\ K)$ $E_a=84.9\ kJ/mol$