where $I=\frac{I_{o}}{r^{2}}$
$\Longrightarrow L=10 \times \log _{10}\left(\frac{1}{r^{2}}\right)$
Now, $160=10 \times \log _{10}\left(\frac{1}{(100)^{2}}\right)$ $.....(1)$
Let the altitude be $x$ m for the sound intensity of $120\, dB.$
So, $120=10 \log _{10}\left(\frac{1}{x^{2}}\right)$ $.....(2)$
$16-12=\log _{10} \frac{1}{10000}-\log _{e} \frac{1}{x^{2}}$
$4=\log _{10} \frac{x^{2}}{10000} \Longrightarrow x=10000 m=10 k m$ above the ground.
