1. Given data

• Reaction: $\mathrm{A} \rightleftharpoons \mathrm{B}$
• Enthalpy of forward reaction: $\Delta H_{\text{forward}} = +20\,\text{kJ mol}^{-1}$
• Activation energy of uncatalysed forward reaction: $E_{a,f}^{(u)} = 300\,\text{kJ mol}^{-1}$
• Temperature of catalysed reaction: $T_1 = 27^\circ\text{C} = 300\,\text{K}$
• Temperature of uncatalysed reaction: $T_2 = 327^\circ\text{C} = 600\,\text{K}$
• Rate of catalysed forward reaction at $300\,\text{K}$ equals rate of uncatalysed forward reaction at $600\,\text{K}$.

Since reactant concentration is same, equal rates imply equal rate constants: $k_c(300) = k_u(600)$

2. Use Arrhenius equation

For the same reaction, assuming the pre-exponential factor remains the same, $k = A e^{-E_a/RT}$

So, $A e^{-E_{a,f}^{(c)}/(R\cdot 300)} = A e^{-300000/(R\cdot 600)}$

Cancelling $A$ and taking exponents equal: $\frac{E_{a,f}^{(c)}}{300} = \frac{300000}{600}$

Be careful with units: easier in kJ mol$^{-1}$: $\frac{E_{a,f}^{(c)}}{300} = \frac{300}{600}$

Thus, $E_{a,f}^{(c)} = 300 \times \frac{300}{600} = 150\,\text{kJ mol}^{-1}$

So the activation energy of catalysed forward reaction is $E_{a,f}^{(c)} = 150\,\text{kJ mol}^{-1}$

3. Relate forward and backward activation energies

For any reaction, $\Delta H = E_{a,f} - E_{a,b}$

Hence, $E_{a,b} = E_{a,f} - \Delta H$

For the catalysed reaction: $E_{a,b}^{(c)} = 150 - 20 = 130\,\text{kJ mol}^{-1}$

4. Final answer

$\boxed{130\,\text{kJ mol}^{-1}}$