Answer
12.42 MeV.
Work Step by Step
See example 30-5.
As in that problem, calculate the binding energy of the last neutron using the masses of a neutron and of the lighter isotope.
$$E_{binding}=\left(m(^{22}_{11}Na)+m(^{1}_{0}n)-m(^{23}_{11}Na)\right)c^2$$
$$ =\left( (21.994437u)+(1.008665u)-(22.989769)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ =12.42MeV$$
