$\Delta H_{rxn} = (E_a)_f - (E_a)_b$. Given $\Delta H_{rxn} = -4.2 \text{ kJ mol}^{-1}$ and $(E_a)_f = 9.6 \text{ kJ mol}^{-1}$. So, $-4.2 = 9.6 - (E_a)_b$, which gives $(E_a)_b = 9.6 + 4.2 = 13.8 \text{ kJ mol}^{-1}$. Since the reaction is exothermic ($\Delta H < 0$), the energy of product $B$ must be lower than reactant $A$. Also, since $(E_a)_f < (E_a)_b$, the forward activation barrier is smaller than the backward activation barrier. This matches graph (2).