Find the 4th term from the end of the G.P.

Here, first term, $a=\frac{2}{27}$

$C$ ommon ratio, $r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3$

Last term, $l=162$

After reversing the given G.P., we get another G.P. whose first term is $l$ and common ratio is $\frac{1}{r}$.

$\therefore 4$ th term from the end $=l\left(\frac{1}{r}\right)^{4-1}=(162)\left(\frac{1}{3}\right)^{3}=6$