$3.01 \times 10^{20}$ molecules are present in =$42 \times 10^{-3}\;g$ of hydrocarbon $(C_xH_y)$

$\therefore 6.023 \times 10^{23}$ molecules are present in

$\qquad= \large\frac{42 \times 10^{-3}}{3.01 \times 10^{20}} $$ \times 6.023 \times 10^{23}$

$\qquad=84$

The molecular formula $=C_6H_{12}$

Hence d is the correct answer.