$$\textbf{Step 1}$$ Chemical reaction for the combustion of ethylene and propylene,
$$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$$
$$C_3H_6+\frac {9}{2}O_2\rightarrow 3CO_2+3H_2O$$
1 mol of $$C_2H_4=28 g=22400\ mL$$
1 mole of $$C_3H_6=42 g=22400\ mL$$
1 mole of $$O_2=32g=22400\ mL$$
$$\textbf{Step 2}$$ Calculation of oxygen require in the combusion of ethylene
As 1 mol ($$22400\ mL$$) of ethylene require 3 mol ($$3\times 22400\ mL$$) of oxygen
$$\therefore 12\ mL$$ ethylene will need $$3\times 12$$ mL $$O_2$$ = $$36\ mL$$ of $$O_2$$
$$\textbf{Step 3}$$ Calculation of oxygen require in the combusion of propylene.
As 1 mol ($$22400\ mL$$) of propylene requires 3 mol ($$3\times 22400\ mL$$) of oxygen
$$\therefore 12\ mL$$ propylene will need $$4.5\times 12$$ mL $$O_2$$ = $$54\ mL$$ $$O_2$$
Therefore option B is correct.