For the reaction $X \rightarrow Y$ the forward activation energy is given as $E_{a,\text{forward}} = 30\ \text{kJ mol}^{-1}$

The energy change of the reaction is $\Delta E = E_Y - E_X = -20\ \text{kJ mol}^{-1}$ Since $\Delta E$ is negative, the reaction is exothermic, so product $Y$ lies $20\ \text{kJ mol}^{-1}$ below reactant $X$.

Let the activated complex have energy $E^\ddagger$. Then, $E_{a,\text{forward}} = E^\ddagger - E_X = 30$

For the reverse reaction $Y \rightarrow X$, $E_{a,\text{reverse}} = E^\ddagger - E_Y$

Using $E_Y = E_X + \Delta E = E_X - 20$ we get $E_{a,\text{reverse}} = E^\ddagger - (E_X - 20)$ $= (E^\ddagger - E_X) + 20$ $= 30 + 20 = 50\ \text{kJ mol}^{-1}$

Therefore, the activation energy for the reverse reaction is $\boxed{50}$