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 \[{{7}^{th}}~term\text{ }{{a}_{7}}~=\text{ }-2187\]