$T_r=T_{(r-1)}+1=10C_{r-1}(\large\frac{x}{3})^{10-(r-1)}(-\large\frac{2}{x^2})^{r-1}$
$\Rightarrow 10C_{r-1}(\large\frac{1}{3})^{n-r} $$(-2)^{r-1} x^{13-3r}$
Let $T_{r+1}$ contains $x^4$
$\therefore 13-3r=4$
(i.e) $r=3$
Hence (B) is the correct answer.