This can be solved through dimensional analysis.
Atomic Masses:
$"Fe = 55.85 g/mol"$
$"O = 16.00 g/mol"$
Formula Mass:
$"Fe"_2"O"_3 ="159.69 g/mol"$
$"Moles of Fe" =73.2cancel("g Fe") xx ("1 mol Fe")/(55.85cancel("g Fe")) = "1.31 mol Fe"$
$"Moles of Fe"_2"O"_3 =1.31 cancel("mol Fe") xx ("2 mol Fe"_2"O"_3)/(4cancel("mol Fe")) = "0.655 mol Fe"_2"O"_3$
$"Mass of Fe"_2"O"_3 = 0.655cancel("mol Fe"_2"O"_3) xx ("159.69 g Fe"_2"O"_3)/(1 cancel("mol Fe"_2"O"_3)) = "104.6 g Fe"_2"O"_3$