${\mathrm{\Delta H}}_{\mathrm{formation} }={\mathrm{IE}}_1+{\mathrm{\Delta Heg}}_1+\mathrm{LE}$

$=495.8+\left(-325\right)+\left(-728.4\right)$

$=-557.6$

$=-5576\times {10}^{-1}\mathrm{KJ}/\mathrm{mol} .$

Note: The above calculation is not for

${\mathrm{\Delta H}}_{\mathrm{formation} }$ but for ${\mathrm{\Delta H}}_{\mathrm{Reaction} }$

But on the basis of given data it is the best ans.