From the question it is given that,
The fourth term of a G.P. is the square of its second term \[=\text{ }{{a}_{4}}~=\text{}{{({{a}_{2}})}^{2}}\]
The first term \[{{a}_{1}}~=\text{ }\text{ }3\]
We know that, \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]
\[\begin{array}{*{35}{l}}
{{a}_{4}}~=\text{ }a{{r}{4\text{ }\text{ }1}} \\ {{a}_{4}}~=\text{ }a{{r}^{3}} \\
{{a}_{2}}~=\text{ }ar \\
\end{array}\]
Now, \[a{{r}^{3}}~=\text{ }{{\left( ar \right)}^{2}}\]
\[\begin{array}{*{35}{l}}
a{{r}^{3}}~=\text{ }{{a}^{2}}{{r}^{2}} \\
{{r}^{3}}/{{r}^{2}}~=\text{ }{{a}^{2}}/a \\
{{r}^{3\text{ }\text{ }2}}~=\text{ }{{a}^{2\text{ }\text{ }1}}~\ldots \text{ }[from\text{}{{a}^{m}}/{{a}^{n}}~=\text{ }{{a}^{m\text{ }\text{ }n}}] \\
r\text{ }=\text{ }a \\
{{a}_{1~}}=\text{ }-3 \\
{{a}_{7}}~=\text{ }a{{r}^{7\text{ }\text{ }1}} \\
{{a}_{7}}~=\text{ }a{{r}^{6}} \\
=\text{ }-3\text{ }\times \text{ }{{\left( -3 \right)}^{6}} \\
=\text{ }-3\text{ }\times \text{ }729 \\
=\text{ }-2187 \\
\end{array}\]
Therefore, the \[{{7}^{th}}~term\text{ }{{a}_{7}}~=\text{ }-2187\]