Answer
0.626 MeV.
Work Step by Step
The reaction is
$$n + \;^{14}_{7}N\rightarrow\;^{14}_{6}C +?$$
Conserve nucleon number: 1 + 14 = 14 + A. We see that A = 1.
Conserve charge: 0 + 7 = 6 + Z. We see that Z = 1.
The unknown product is a proton.
$$n + \;^{14}_{7}N\rightarrow\;^{14}_{6}C +p$$
Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left( m_{n}+m(^{14}_{7}N)-m(^{14}_{6}C)-m_{p} \right)c^2$$
$$ =\left(1.008665u+14.003074u-14.003242u-1.007825u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=0.626MeV$$