Answer
$ r_1=1cm$
Work Step by Step
As $I=\frac{P}{A}$
but for an isotropic point source $A_{sphere}=4{\pi}r^2$
Now let us compare the two intensities $I_1$ and $I_2$ at distance $r_1$ and $r_2$ respectively.
Mathematically, $\frac{I_2}{I_1}=\frac{\frac{P}{4{\pi}r_2^2}}{\frac{P}{4{\pi}r_1^2}}$
After simplifying we get
$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$
we know that intensity ratio between barely audible and painful threshold is given as
$10^{-12}=\frac{I_2}{I_1}$
and $r_2=10Km=10\times1000m=10000m$
put the values
$10^{-12}=\frac{r_1^2}{(10000)^2}$
or $10^{-12}(10000)^2=r_1^2$
taking square root on both sides and arranging the above equation,we get
$r_1=(10000)\sqrt{10^{-12}}=0.01m=1cm$
thus $ r_1=1cm$
