The mass of oxygen required for the rusting of $4.2 \mathrm{g}$ of iron is $(\mathrm{Fe}=56)$

$4\mathrm{Fe}+3{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}\to 4\mathrm{Fe}{\left(\mathrm{OH}\right)}_3$

4 mole of $\mathrm{Fe}$ combines with 3 moles of $O_2$

Now, $4\times 56 \mathrm{g}$ of $\mathrm{Fe}$ combines with $3\times 32 \mathrm{g}$ moles of $O_2$. (Molar mass of iron is $56 \mathrm{g}$ and that of oxygen is 32 )$4.2\mathrm{g} \mathrm{of} \mathrm{Fe} \mathrm{combines} \mathrm{with} = 4.2\times 3\times 32/4\times 56=1.8\mathrm{g}$