Solution:

$n^{th}$ 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 $n^{th}$ term
Given that, last term, $I=1 / 4374$
$\begin{array}{l}
r=t_{2} / t_{1}=(1 / 6) /(1 / 2) \\
=1 / 6 \times 2 / 1 \\
=1 / 3 \\
n=4
\end{array}$
Therefore, $a_{n}=\mid(1 / r)^{n-1}$
$\begin{array}{l}
a_{4}=1 / 4374(1 /(1 / 3))^{4-1} \\
=1 / 4374(3 / 1)^{3} \\
=1 / 4374 \times 3^{3} \\
=1 / 4374 \times 27 \\
=1 / 162
\end{array}$
Hence, $4^{\text {th }}$ term from last is $1 / 162$.