#### Answer

At a distance of 50 m, the sound intensity level will be 80 dB.

#### Work Step by Step

Let the sound intensity at a distance of 5.0 meters be $I_1$. For every 10 dB change, the intensity changes by a factor of 10. Since the decibel level decreases by 20, the intensity level decreases by a factor of 100. Therefore, $I_2 = \frac{I_1}{100}$.
We can find the distance $r_2$ where $I_2 = \frac{I_1}{100}$:
$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$
$r_2^2 = \frac{I_1~r_1^2}{I_2}$
$r_2 = \sqrt{\frac{I_1}{I_2}}~r_1$
$r_2 = \sqrt{\frac{I_1}{(I_1/100)}}~r_1$
$r_2 = \sqrt{100}~(5.0~m)$
$r_2 = 50~m$
At a distance of 50 m, the sound intensity level will be 80 dB.