Answer
The new intensity becomes $\frac{1}{4}$ of initial intensity.
Work Step by Step
Let intensity be $I$ and distance be $R$.
Therefore,
$\begin{align}
& I\propto \frac{1}{{{R}^{2}}} \\
& I=\frac{k}{{{R}^{2}}}
\end{align}$
Now, when the distance of the source doubles, then ${{R}_{1}}=2R$.
The new intensity ${{I}_{1}}$ becomes
$\begin{align}
& {{I}_{1}}=\frac{k}{R_{1}^{2}} \\
& {{I}_{1}}=\frac{k}{{{\left( 2R \right)}^{2}}} \\
& {{I}_{1}}=\frac{k}{4R} \\
& {{I}_{1}}=\frac{1}{4}I
\end{align}$
