First we need a balanced reaction equation:

4 Fe + 3 O_{2} →2 Fe_{2}O_{3}

Next, since 50.0 grams of iron(III) oxide is only 78.7% of the total mass of iron(III) oxide that the reaction could have produced we need to know what mass 100.0% would have been, so we setup the ratio:

50.0g/x = 78.7%/100% , which rearranges algebraically to get x alone to: 50g*100%/78.7% , the percent "units" cancel out and we get 63.5324015 grams of iron(III) oxide [alternatively we could have turned the 78.7% into decimal form and solved for x in the equation 50.0/x=0.787].

Now we know what mass the reaction should have produced, so we can setup a stoichiometry (mole) ratio and use the molar masses given [g⋅mol^{-1} is the same thing as g/mol which can be flipped to mol/g if necessary, just keep in mind the that denominator is technically (invisible) 1 mole] to solve for the initial mass of iron needed. The mole ratio used will be the coefficients from the above reaction equation 4 mole Fe = 2 mole Fe_{2}O_{3}.

63.5324015 g Fe_{2}O_{3} × 1 mol Fe_{2}O_{3}/159.70 g Fe_{2}O_{3} × 4 mol Fe / 2 mol Fe_{2}O_{3} × 55.85 g Fe / 1 mol Fe = 44.4368769 g Fe ≈ 44.4 g Fe

Since the values in the problem all go to the tenths place (the molar masses are values that we looked up, not measured, so their precision to the hundredths place gets ignored [plus the have an extra digit of significance, and we always let the least amount of precision limit our answers]) then our final answer gets rounded to the tenths place too.