(d)
$E_{a}(A \rightarrow B) = 80\ kJ\ mol^{- 1}$
Heat of reaction $(A \rightarrow B) = 200\ kJ\ mol^{- 1}$
For $(B \rightarrow A)\ backward\ reaction,$
$E_{a}(B \rightarrow A) = E_{a}(A \rightarrow B) + heat\ of\ reaction$
$= 80 + 200 = 280$kJ $mol^{- 1}$