Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110 kJ mol–1,

$\Delta_{r} H$ for a reaction is defined as the difference between $\Delta_{f} H$ value of products and $\Delta_{H} H$ value of reactants.

$\Delta, H=\sum \Delta_{f} H$ (products) $-\sum \Delta_{f} H$ (reactants)

For the given reaction,

$\mathrm{N}_{2} \mathrm{O}_{4(g)}+3 \mathrm{CO}_{(g)} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{(g)}+3 \mathrm{CO}_{2(g)}$

$\Delta_{r} H=\left[\left\{\Delta_{f} H\left(\mathrm{~N}_{2} \mathrm{O}\right)+3 \Delta_{f} H\left(\mathrm{CO}_{2}\right)\right\}-\left\{\Delta_{f} H\left(\mathrm{~N}_{2} \mathrm{O}_{4}\right)+3 \Delta_{f} H(\mathrm{CO})\right\}\right]$

Substituting the values of $\Delta_{f} H$ for $\mathrm{N}_{2} \mathrm{O}, \mathrm{CO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}$, and $\mathrm{CO}$ from the question, we get:

$\Delta_{r} H=\left[\left\{81 \mathrm{~kJ} \mathrm{~mol}^{-1}+3(-393) \mathrm{kJ} \mathrm{mol}^{-1}\right\}-\left\{9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}+3(-110) \mathrm{kJ} \mathrm{mol}^{-1}\right\}\right]$

$\Delta_{r} H=-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, the value of $\Delta_{r} H$ for the reaction is $-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$.