In the expansion of ${(3x-\frac{1}{x^2})}^{10}$, the 5^th term from the end is

r^th term from end $={(n-r+2)}^{\text{th}}$ term from beginning

5^th term from end $={(10-5+2)}^{\text{th}}$ term from beginning

$=7^{\text{th}}$ term from beginning

$T_7={}^{10}{\text{C}}_6{\left(3x\right)}^{10-6}{\left(\frac{-1}{x^2}\right)}^6$

$={}^{10}{\text{C}}_6{3}^4.x^4.\frac{1}{x^{12}}{x}^8$

$=\frac{{}^{10}{\text{C}}_6{.3}^4}{x^8}$

$=\frac{17010}{x^8}$