Answer
$-44.35\,kJ/mol$
Work Step by Step
$\Delta H_{rxn}^{\circ}=\Sigma n\Delta H_{f}^{\circ}(\text{products})-\Sigma m\Delta H_{f}^{\circ}(\text{reactants})$
where $m$ and $n$ denote the stoichiometric coefficients for the reactants and products.
Then,
$\Delta H_{rxn}^{\circ}=78.67\, kJ/mol=$
$[\Delta H_{f}^{\circ}(AgNO_{2}(s))+\frac{1}{2}\Delta H_{f}^{\circ}(O_{2}(g))]-[\Delta H_{f}^{\circ}(AgNO_{3}(s))]=$
$[\Delta H_{f}^{\circ}(AgNO_{2}(s))+\frac{1}{2}\times0]-(-123.02\,kJ/mol)=$
$\Delta H_{f}^{\circ}(AgNO_{2}(s))+123.02\,kJ/mol$
$\implies \Delta H_{f}^{\circ}(AgNO_{2}(s))=(78.67-123.02)kJ/mol$
$=-44.35\,kJ/mol$
