Answer
$7.700\times10^{8}$ kJ/mol nucleons
Work Step by Step
Mass defect $\Delta m=[m_{p}Z+(A-Z)m_{n}]-M$
$=[1.00783\,g/mol\times8+(16-8)1.00867\,g/mol]-15.99492\,g/mol$$=0.13708\,g/mol=1.3708\times10^{-4}\,kg/mol$
Binding energy per mol $E_{b}=(\Delta m)c^{2}=(1.3708\times10^{-4}\,kg/mol)(2.99792\times10^{8}\,m/s^{2})^{2}$$=1.23201\times10^{13}\,J/mol$
Binding energy per mole of nucleons
$\frac{E_{b}}{n}=\frac{1.23201\times10^{13}\,J/mol}{16\,nucleons}=7.700\times10^{11}$ J/mol nucleons
$=7.700\times10^{8}$ kJ/mol nucleons