The middle term in the expansion of

(b) 252

Here, $n$, i. e., 10 , is an even number.

$\therefore$ Middle term $=\left(\frac{10}{2}+1\right)$ th term $=6$ th term

Thus, we have

$T_{6}=T_{5+1}$

$={ }^{10} C_{5}\left(\frac{2 x^{2}}{3}\right)^{10-5}\left(\frac{3}{2 x^{2}}\right)^{5}$

$=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times \frac{2^{5}}{3^{5}} \times \frac{3^{5}}{2^{5}}$

$=252$