Solution:

The $n^{th}$ term from the end is given by:
$a_{n}=I(1 / r)^{n-1}$ where, $I$ is the last term, $r$ is the common ratio, $n$ is the nth term
Given that, last term, $\mid=162$
$\mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=(2 / 9) /(2 / 27)$
$=2 / 9 \times 27 / 2$
$=3$
$n=4$
Therefore, $a_{n}=I(1 / r)^{n-1}$
$a_{4}=162(1 / 3)^{4-1}$
$=162(1 / 3)^{3}$
$=162 \times 1 / 27$
$=6$
Therefore, $4^{\text {th }}$ term from last is $6 .$