The middle term in the expansion ${\left(3x-\frac{{\displaystyle 2}}{{\displaystyle x^2}}\right)}^{10}$ is

$n=10=$ even

The middle term $=\frac{n}{2}+1=5+1=$ 6^th term

6^th term $\left(t_6\right)=t_{5+1}={}^{10}{C}_5{\left(3x\right)}^{10-3}{\left(\frac{-2}{x^2}\right)}^5$

$=\frac{10!}{(10-5)!5!}{\left(3x\right)}^5\left(\frac{{\displaystyle -32}}{{\displaystyle x^{10}}}\right)$

$=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\left(32\right)\left(x^{5-10}\right)(-32)$

$\begin{array}{l}=6\times 7\times 6(-1024)\left(x^{-5}\right)\\ =\frac{258048}{x^5}\end{array}$.