The answer is -1839.5 kJ/mol. This was calculated using the given formula for ΔH. You have to calculate the sum of ΔHf for products and subtract from it the ΔHf of the reactants.
ΔHrxn = (2 mol *-1699.8 kJ/mol) - (3 mol * -520 kJ/mol) where 2 mol Al2O3 and 3 mol MnO2 came from the given reaction. The other reactant and products are in their standard form so their ΔHf is zero.