Answer
four times less than the initial sound intensity
Work Step by Step
Let $I$ be the sound intensity, d the distance,
$I$ varies inversely as $d^{2},\ \displaystyle \qquad I=\frac{k}{d^{2}}$
When the distance $d$ changes to $2d$ (twice as far),
the new intensity is
$I_{2}=\displaystyle \frac{k}{(2d)^{2}}=\frac{1}{4}(\frac{k}{d^{2}})=\frac{1}{4}I$,
that is, the new intensity is
four times less than the initial sound intensity
