$T_3=nC_2 (x^2)^{n-2} (-\large\frac{1}{4})^2$
$\quad=\large\frac{n(n-1)}{32}$$ x^{2n-4}$
Coefficient of $T_3=\large\frac{n(n-1)}{32}$
$\therefore \large\frac{n(n-1)}{32}=$$31$
$\Rightarrow n(n-1)=32(32-1)$
$\Rightarrow n=32$
Hence (C) is the correct answer.