Answer
218.2 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,
$\Delta H_{rxn}^{\circ}=436.4\,kJ/mol=$
$[\Delta H_{f}^{\circ}(H)+\Delta H_{f}^{\circ}(H)]-\Delta H_{f}^{\circ}(H_{2})$
$=2\Delta H_{f}^{\circ}(H)-0=2\Delta H_{f}^{\circ}(H)$
$\implies \Delta H_{f}^{\circ}(H)=\frac{436.4\,kJ/mol}{2}=218.2\,kJ/mol$