The correct option is
B −242.3 kcal/mol
Given, 2Fe+32O2→Fe2O3....(i) △G=−177 KCal
4Fe2O3+Fe→3Fe3O4....(ii) △G=−19 KCal
Divide by (iii) in reaction (ii)
43Fe2O3+13Fe→Fe3O4....(iii) △G=−193 KCal
Similarly, multiply by 43 in reaction (i)
83Fe+43×32O2→43Fe2O3....(iv) △G=−177×43 Kcal
Now from (iii) and (iv)
83Fe+2O2→43Fe2O3....(iv) −177×43 Kcal
43Fe2O3+13Fe2→Fe3O4....(iii) −193 Kcal
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Now adding (iv) and (iii)
3Fe(s)+2O2(g)→Fe3O4 △G=(−177×43−193) Kcal
△G=−242.3 Kcal/mol