$ \Delta H_f\: of\: CH_4 = -74.8\: kJ/mol\: \Delta H_f\: of\: H = \large\frac{435.4}{2}$$ = 217.7 \: kJ/mol$
$\Delta H_f\: of \: C = 718.4\: kJ/mol$
Now, $ \Delta H = \sum \Delta H_f^{\circ}(products) - \sum \Delta H_f^{\circ}o(reactants)$
$ = (4\times217.7 + 718.4) – (-74.8) = 1664\: kJ$
$ \Delta H_{C-H} = \large\frac{Total\: energy \: required}{No.\: of\: C-H \: broken }$$ = \large\frac{1664}{4}$$ = 416\: kJ/mol$
Ans : (C)