Mol.wt.of mixture of $NO_2$ and $N_2O_4=38.3\times 2$
$\Rightarrow 76.6$
Let $a$ g of $NO_2$ present in 100g mixture.
Mole of $NO_2$+mole of $N_2O_4$=Mole of mixture
$\large\frac{a}{46}+\frac{100-a}{92}=\frac{100}{76.6}$
$a=20.10g$
$\therefore$ Mole of $NO_2$ in mixture =$\large\frac{20.10}{46}$
$\Rightarrow 0.437$
Hence (c) is the correct answer.