Answer
See the detailed answer below.
Work Step by Step
The needed energy to remove one neutron from $\rm ^4_2He$. So first we need to find the difference between the mass of $\rm ^4_2He$ and $\rm ^3_2He+n$.
Thus,
$$\Delta E=mc^2=(m_{^3_2He}-[m_{^4_2He}+m_n])c^2$$
Plugging from Appendix B;
4.002603
$$\Delta E=\left[ (4.002603\times 1.67\times 10^{-27})-(3.016029\times 1.67\times 10^{-27})-( 1.008665\times 1.67\times 10^{-27}) \right](3\times 10^8)^2$$
$$\Delta E_1=\color{red}{\bf 20.7517}\;\rm MeV$$
Now we need to repeat this process for $\rm ^{13}_6C$;
$$\Delta E=mc^2=(m_{^{13}_6C}-[m_{^{12}_6C}+m_n])c^2$$
$$\Delta E_2=\left[ (13.003355\times 1.67\times 10^{-27})-(12.000000\times 1.67\times 10^{-27})-( 1.008665\times 1.67\times 10^{-27}) \right](3\times 10^8)^2$$
$$\Delta E_2=\color{red}{\bf 4.98808}\;\rm MeV$$
To find how much $E_2$ is greater than $E_1$;
$$\dfrac{E_1}{E_2}=\dfrac{20.7517}{4.98808}=4.16$$
$$E_1=\color{red}{\bf4.16} E_2$$
