$\begin{align}  & \beta =(10dB)\log \frac{I}{{{I}_{\circ }}}\Rightarrow {{\beta }_{2}}-{{\beta }_{1}}=(10dB)\log \frac{{{I}_{2}}}{{{I}_{1}}}=10\log {{(\frac{{{r}_{1}}}{{{r}_{2}}})}^{2}}\Rightarrow -12=10\log {{(\frac{5}{{{r}_{2}}})}^{2}}=20\log \frac{5}{{{r}_{2}}} \\  & \Rightarrow \log \frac{{{r}_{2}}}{5}=0/6=2\log 2=\log {{2}^{2}}\Rightarrow \frac{{{r}_{2}}}{5}=4\Rightarrow {{r}_{2}}=20m\Rightarrow bayad\,15m\,\,dortar\,shavad \\ \end{align}$