Explanation

(c)
$General\ term\ T_{r + 1} = \ ^{10}C_{r}\left( \frac{x}{2} \right)^{10 - r}\left( - \frac{3}{x^{2}} \right)^{r}$
$= \ ^{10}C_{r} \bullet \frac{x^{10 - 3r} \bullet ( - 1)^{r} \bullet 3^{r}}{2^{10 - r}}$
For the coefficient of $x^{4}$put
$10 - 3r = 4$
$\Longrightarrow \ \ \ \ \ \ r = 2$
Hence, coefficient of $x^{4}\$is
$\ \ ^{10}C_{2} \bullet \frac{3^{2}}{2^{8}} = \frac{405}{256}$