${\mathrm{CH}}_{4\left(9\right)}+2{\mathrm{O}}_{2\left(g\right)}⟶{\mathrm{CO}}_{2\left(g\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(i\right)}$

$\Delta \mathrm{H}=-200\mathrm{ }\mathrm{K}$.Cal, now $\Delta \mathrm{E}$ for the same process at $300\mathrm{ }\mathrm{K}$ (in K.Cal)

According to first law of Thermodynamics expression for the relation between enthalpy and internal energy is $\Delta H=\Delta E+nRT$ also written as $\Delta E=\Delta H-nRT$

 Given $\Delta H=-200;T=300K;R=2$

 $\Delta n_g=$ no. of mole of gas products - no. of mole of gas reactants

 $\Delta n_g=1-3=-2$

 $\Delta E=-200-\left(-2\times 2\times 300\times {10}^{-3}\right).{10}^{-3}$ because value asked in Kcal.

 $\Delta E=-198.8$

Hence the correct option is (D).