$NO$ reacts with $O_2$ to form ${NO}_2$. When $10 g$ of ${NO}_2$ is formed during the reaction, the mass of $O_2$ consumed is

Balanced chemical reaction

$NO+\frac{1}{2}{O}_2\to {NO}_2.$

0.5 moles of ${\mathrm{O}}_2 \mathrm{is} \mathrm{used} \mathrm{to} \mathrm{form}1\mathrm{mole} \mathrm{of} {\mathrm{NO}}_2$. 

Moles of ${NO}_{2}formed=10/46=0.217moles$.

Moles of $O_{2}consumed=0.5\times 0.217=0.1087\text{ moles }$.

Mass of $O_{2}consumed=0.1087\times 32=3.48g$.

Therefore, option C is correct.