Answer
$-3923.9\,kJ/mol$
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,
$\Delta H_{rxn}^{\circ}=$
$[6\Delta H_{f}^{\circ}(CO_{2}(g))+6\Delta H_{f}^{\circ}(H_{2}O(l))]-[\Delta H_{f}^{\circ}(C_{6}H_{12}(l))+9\Delta H_{f}^{\circ}(O_{2}(g))]$
$=[(6\times-393.5\,kJ/mol)+(6\times-285.8\,kJ/mol)]-[(-151.9\,kJ/mol)+(9\times0)]$
$=-3923.9\,kJ/mol$
