Answer
When $d$ is doubled, the intensity is four times smaller.
Work Step by Step
$I=\frac{k}{d^2}$
When $d$ is doubled,
$\frac{k}{(2d)^2}=\frac{k}{4d^2}$
Since $\frac{k}{d^2}=I$,
$\frac{k}{4d^2}=\frac{I}{4}$
When $d$ is doubled, the intensity is four times smaller.
