From Arrhenius equation, $$\log k = \log A - {{{E_a}} \over {2.303RT}}$$

$$\therefore$$ $$\log {k_{500}} = \log A - {{{E_{{a_1}}}} \over {2.303R{T_1}}}$$, $$\log {k_{400}} = \log A - {{{E_{{a_2}}}} \over {2.303R{T_2}}}$$

Given, $${k_{500}} = {k_{400}}$$ $$\therefore$$ $$\log {k_{500}} = \log {k_{400}}$$

or, $${{{E_{{a_1}}}} \over {{T_1}}} = {{{E_{{a_2}}}} \over {{T_2}}}$$ or, $${{{E_{{a_1}}}} \over {500}} = {{{E_{{a_2}}}} \over {400}}$$ or, $${{{E_{{a_1}}}} \over {{E_{{a_2}}}}} = {{500} \over {400}} = {5 \over 4}$$ ...... [1]

According to given data, $${E_{{a_1}}} - {E_{{a_2}}} = 20$$

$$\therefore$$ Substituting in equation [1], we get $${{{E_{{a_1}}}} \over {{E_{{a_1}}} - 20}} = {5 \over 4}$$

or, $$4{E_{{a_1}}} - 5{E_{{a_1}}} - 100$$ or, $${E_{{a_1}}} = 100$$ kJ mol^$$-$$1

$$\therefore$$ $${E_{{a_2}}} = 100 - 20 = 80$$ kJ mol^$$-$$1