Consider an endothermic reaction, $$X \longrightarrow Y$$ with the activation energies $$E_b$$ and $$E_f$$ for the backward and forward reactions, respectively. In general

For any elementary reversible reaction $$X \rightleftharpoons Y$$ the potential-energy profile has a single transition state (activated complex) located at energy $$E_{\ddagger}$$ above the reactants.

Define the two activation energies as
  • forward direction : $$E_f = E_{\ddagger} - E_X$$
  • backward direction : $$E_b = E_{\ddagger} - E_Y$$
where $$E_X$$ and $$E_Y$$ are the potential energies of $$X$$ and $$Y$$, respectively.

The heat (enthalpy) of reaction is the difference between product and reactant energy:

$$\Delta H = E_Y - E_X \qquad -(1)$$

Combine the two definitions to relate the activation energies:

$$E_f - E_b = (E_{\ddagger} - E_X) - (E_{\ddagger} - E_Y) = E_Y - E_X = \Delta H \qquad -(2)$$

For an endothermic reaction, $$\Delta H \gt 0$$ (products are at higher energy than reactants). Equation $$(2)$$ then gives

$$E_f - E_b = \Delta H \gt 0 \quad\Longrightarrow\quad E_f \gt E_b$$

Thus the activation energy for the backward reaction is smaller than that for the forward reaction.

Option A which is: $$E_b \lt E_f$$