b
According to arrhenius equation

\(\log \,\frac{{{k_2}}}{{{k_1}}}\, = \,\frac{{{E_a}}}{{2.303\,\,R}}\,\left( {\frac{1}{{{T_1}}}\, - \,\frac{1}{{{T_2}}}} \right)\)

\(\log \,\frac{{1.3 \times {{10}^{ - 3}}}}{{1.3 \times {{10}^{ - 4}}}}\, = \,\frac{{{E_a}}}{{2.303 \times 8.314}}\,\left[ {\frac{1}{{373}} - \frac{1}{{423}}} \right]\)

\(1 = \frac{{{E_a}}}{{2.303 \times 8.314}}\,\left[ {\frac{1}{{373}} - \frac{1}{{423}}} \right]\)

\({E_a}\, = \,60\,\,kJ/mole\)