Given, $r_{1}=8 \times 10^{3} \,m , L_{1}=30 \,dB$

$r_{2}=80 \,m , L_{2}=?$

As, $L=10 \log _{10}\left(\frac{I}{I_{0}}\right)$

Also, $\quad I=\frac{P}{A}=\frac{P}{4 \pi r^{2}} \Rightarrow I \propto \frac{1}{r^{2}}$

$\therefore L_{2}-L_{1} =10 \log _{10}\left(\frac{I_{2}}{I_{0}}\right)-10 \log _{10}\left(\frac{I_{1}}{I_{0}}\right)$ $=10 \log _{10}\left(\frac{I_{2}}{I_{1}}\right)$ $=10 \log _{10}\left(\frac{r_{1}}{r_{2}}\right)^{2}$

$L_{2}-30=10 \log _{10}\left(\frac{8000}{80}\right)^{2}$ $=20 \times 2=40$ $\Rightarrow \quad L_{2} =70 \,dB$