According to the given question,

G.P. \[1/27,\text{ }1/9,\text{ }1/3,\text{ }\ldots \ldots ,\text{ }81\]

Here, \[a\text{ }=\text{ }1/27,\]

\[common\text{ }ratio\text{ }\left( r \right)\text{ }=\text{ }\left( 1/9 \right)/\text{ }\left( 1/27 \right)\text{ }=\text{ }3\]

And

\[l\text{ }=\text{ }81\]

We know that,

\[l\text{ }=\text{ }{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}~=\text{ }81\]

\[\left( 1/27 \right){{\left( 3 \right)}^{n\text{ }-\text{ }1}}~=\text{ }81\]

\[{{3}^{n\text{ }-\text{ }1}}~=\text{ }81\text{ }x\text{ }27\text{ }=\text{ }2187\]

\[{{3}^{n\text{ }-\text{ }1}}~=\text{ }{{3}^{7}}\]

\[n\text{ }-\text{ }1\text{ }=\text{ }7\]

\[n\text{ }=\text{ }8\]

Hence, there are \[8\text{ }terms\] the given G.P.

\[{{4}^{th}}~term\] from the beginning is \[{{t}_{4}}~and\text{ }the\text{ }{{4}^{th}}~term~\] from the end is \[\left( 8\text{ }-\text{ }4\text{ }+\text{ }1 \right)\text{ }=\text{ }{{5}^{th}}~term\text{ }({{t}_{5}})\]

Therefore,

The product of \[{{t}_{4~}}and\text{ }{{t}_{5}}\] 

\[=\text{ }a{{r}^{4\text{ }-\text{ }1}}~x\text{ }a{{r}^{5\text{ }-\text{ }1}}~=\text{ }a{{r}^{3}}~x\text{ }a{{r}^{4}}\]

 \[=\text{ }{{a}^{2}}{{r}^{7}}~=\text{ }{{\left( 1/27 \right)}^{2}}{{\left( 3 \right)}^{7}}~=\text{ }3\]