Explanation

(b) $I \propto \frac{1}{r^{2}} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{2^{2}}{(40)^{2}} = \frac{1}{400} \Rightarrow I_{1} = 400I_{2}$
Intensity level at point $1,\ L_{1} = 10\log_{10}\left( \frac{I_{1}}{I_{0}} \right)$
and intensity at point $2,\ L_{2} = 10\log_{10}\left( \frac{I_{2}}{I_{0}} \right)$
$\therefore L_{1} - L_{2} = 10\log{\frac{I_{1}}{I_{2}} = 10\log_{10}{(400)}}$
$\Rightarrow L_{1} - L_{2} = 10 \times 2.602 = 26$
$L_{2} = L_{1} - 26 = 80 - 26 = 54\ dB$