Based on the value of B.E. Given, 

${∆}_{\mathrm{f}}{\mathrm{H}}^0 \mathrm{of} {\mathrm{N}}_2{\mathrm{H}}_{4\left(\mathrm{g}\right)}$ is: Given BE of : N-N is 159 kJ mol^-1; H-H is 436 kJ mol^-1; N=N is 941 kJ mol^-1; N-H is 398 kJ mol^-1

$\begin{array}{r}\Rightarrow \mathrm{△}\mathrm{H}\left(\text{ bond breaking }\right)=\mathrm{\Delta H}(\mathrm{N}\equiv )+2\mathrm{\Delta H}(\mathrm{H}-\mathrm{H})\\ =941+2\left(436\right)\\ =1813\end{array}$

$\begin{array}{r}\mathrm{\Delta H}\left(\text{ bond formation }\right)=-\left[\mathrm{\Delta H}\right(\mathrm{N}-\mathrm{N})+4\mathrm{\Delta H}(\mathrm{N}-\mathrm{H}\left)\right]\\ =-[159+4(398\left)\right]\\ =-1751\end{array}$

$\begin{array}{r}\Rightarrow \mathrm{\Delta H}=\mathrm{\Delta H}\left(\text{ bond formation }\right)+\mathrm{\Delta H}\left(\text{ bond breaking }\right)\\ =-1751+1813\\ =62\mathrm{kJ}/\mathrm{mol}\end{array}$