Answer
The sound intensity level at a distance of 1.0 km is 109.5 dB
Work Step by Step
Let $I_2$ be the intensity at a distance of $r_2 = 1.0~km$. We can find an expression for $I_2$ in terms of $I_1$.
$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$
$I_2 = \frac{r_1^2}{r_2^2}~I_1$
$I_2 = \frac{(30~m)^2}{(1000~m)^2}$
$I_2 = (9\times 10^{-4})~I_1$
We can find the sound intensity level $\beta_2$.
$\beta_2 = 10~log(\frac{I_2}{I_0})$
$\beta_2 = 10~log(\frac{(9\times 10^{-4})~I_1}{I_0})$
$\beta_2 = 10~log(9\times 10^{-4})+10~log(\frac{I_1}{I_0})$
$\beta_2 = (-30.5~dB)+140~dB$
$\beta_2 = 109.5~dB$
The sound intensity level at a distance of 1.0 km is 109.5 dB.
