The term independent of $x$  in the expansion of ${\left(x-\frac{3}{x^2}\right)}^{18}$  is  

Given  expansion  ${\left(x-\frac{3}{x^2}\right)}^{18}$

 We know that $r^{th}$  term in the expansion of ${(x+a)}^n$ is

$\begin{array}{c}{T}_{r+1}{=}^n{\text{C}}_r{x}^{n-r}{a}^r\\ \end{array}$

then

  $T_{r+1}={}^{18}{\text{C}}_r{x}^{18-r}{\left(-\frac{3}{x^2}\right)}^r$

$T_{r+1}={}^{18}{\text{C}}_r{x}^{18-r}{\left(-3\right)}^r{\left(x^{-2}\right)}^r$

       $={}^{18}{\text{C}}_r{(-3)}^r{x}^{18-3r}\mathrm{...................}\left(1\right)$  

And this will be

Without $x$  only if $x^{18-3r}$ compare with $x^0$  term independent of $x$  in the expansion

$\Rightarrow x^{18-3r}=x^0$

$\Rightarrow 18-3r=0$

 $\because r=6$  in Eqn (1)

We get

$\begin{array}{c}{}^{18}{\text{C}}_6{\left(-3\right)}^6{x}^{18-18}\\ \end{array}$

$\begin{array}{c}\because {}^{18}{\text{C}}_6{3}^6\\ \end{array}$