Answer
$\Delta H^{\circ}_{f}=-1265.6 \text{ kJ/mol}$
Work Step by Step
$\Delta H^{\circ}_{rxn} = H^{\circ}_{f} final-H^{\circ}_{f}initial$
Here since the $\Delta H^{\circ}_{rxn}$ value is given and $H^{\circ}_{f} final$ can be taking the sum of $\Delta H^{\circ}_{f}$ of $CO_{2}$ and $H_{2}O$, $H^{\circ}_{f}initial$ can be found as,
$\ H^{\circ}_{f}initial = H^{\circ}_{f} final-\Delta H^{\circ}_{rxn} = (-4075.8 kJ/mol) - (-2810.2 kJ/mol)
= -1265.6 kJ/mol$
Since the formation enthalpy of $O_{2}(g)$ is 0, $\ H^{\circ}_{f}initial = H^{\circ}_{f} fructose$
