The energy needed to remove one neutron from a nucleus is its neutron separation energy…

Step 1. Carbon-13. $M_\text{left} = 12.000000$ u. So $12.000000\ \text{u} + 1.008665\ \text{u} - 13.003355\ \text{u} = 0.005310$ u, and $S = 0.005310\ \text{u} \times 931.494\ \text{MeV/u} = 4.946$ MeV.

Step 2. Carbon-12. $M_\text{left} = 11.011434$ u. So $11.011434\ \text{u} + 1.008665\ \text{u} - 12.000000\ \text{u} = 0.020099$ u, and $S = 0.020099\ \text{u} \times 931.494\ \text{MeV/u} = 18.72$ MeV.

Why. Carbon-12 has $6$ neutrons, an even number, in complete pairs. Carbon-13 has $7$, and the last one has no partner. Nucleons bind more tightly when they pair up, so the unpaired neutron is held loosely. This pairing is one reason why nuclei with even numbers of protons and neutrons are the most stable.