$\mathrm{\Delta H}$ for $3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{O}}_3\left(\mathrm{g}\right)$ is $+272\mathrm{ }\mathrm{k}\mathrm{J}$

Then $\mathrm{\Delta H}$ for ${\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{O}}_3\left(\mathrm{g}\right)\to 2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)$ is $\left[{\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}=-32\mathrm{kJ}\right]$

For the formation of ${\mathrm{Ag}}_2\mathrm{O}$,

$2\mathrm{Ag}+1/2{\mathrm{O}}_2\to {\mathrm{Ag}}_2\mathrm{O} {\Delta }_f\mathrm{H}=-32 \mathrm{kJ}$

On multiplying the above reaction by 2 and reversing it, we get,

$\begin{array}{l}2{\mathrm{Ag}}_2\mathrm{O}\left(s\right)\to 4\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(g\right)\Delta {\mathrm{H}}_1=+64 \mathrm{kJ}\\ 3{\mathrm{O}}_2\left(g\right)\to 2{\mathrm{O}}_3\left(g\right)\Delta H_2=+272 \mathrm{kJ}\end{array}$

On subtracting above two reactions we get,

$2{\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+2{\mathrm{O}}_3\left(\mathrm{g}\right)\to 4\mathrm{Ag}\left(\mathrm{s}\right)+4{\mathrm{O}}_2\left(\mathrm{g}\right) \mathrm{\Delta H}={\mathrm{\Delta H}}_1-{\mathrm{\Delta H}}_2$

On dividing the above reaction by 2 , we get

$$\begin{gathered} {\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{O}}_3\left(\mathrm{g}\right)\to 2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right) \\ \mathrm{\Delta H}=\frac{{\mathrm{\Delta H}}_1-{\mathrm{\Delta H}}_2}{2} \end{gathered}$$

Thus,

$\begin{array}{l}\Rightarrow \mathrm{\Delta H}=\frac{64-272}{2}\\ \Rightarrow \mathrm{\Delta H}=-104 \mathrm{kJ}\end{array}$

Hence, option (C) is correct.