Answer
$r_f = 0.675~m$
Work Step by Step
Let $I_i$ be the initial sound intensity.
We can find an expression for $I_i$:
$\beta_i = (10~dB)~log~\frac{I_i}{I_0}$
$\frac{\beta_i}{10} = log~\frac{I_i}{I_0}$
$I_i = 10^{\frac{\beta_i}{10}}~\cdot I_0$
Let $I_f$ be the final sound intensity.
We can find an expression for $I_f$:
$\beta_1+5~dB = (10~dB)~log~\frac{I_f}{I_0}$
$\frac{\beta_1}{10}+0.5 = log~\frac{I_f}{I_0}$
$10^{(\frac{\beta_1}{10}+0.5)} = \frac{I_f}{I_0}$
$10^{0.5}~\cdot 10^{\frac{\beta_1}{10}} = \frac{I_f}{I_0}$
$I_f = 10^{0.5}~\cdot 10^{\frac{\beta_1}{10}}~\cdot I_0$
We can find $r_f$:
$Power = I_i~4\pi~r_i^2 = I_f~4\pi~r_f^2$
$r_f^2 = \frac{I_i~4\pi~r_i^2}{I_f~4\pi}$
$r_f^2 = \frac{I_i~r_i^2}{I_f}$
$r_f = \sqrt{\frac{I_i}{I_f}}~\cdot r_i$
$r_f = \sqrt{\frac{10^{\frac{\beta_i}{10}}~\cdot I_0}{10^{0.5}~\cdot 10^{\frac{\beta_1}{10}}~\cdot I_0}}~\cdot (1.20~m)$
$r_f = \sqrt{\frac{1}{10^{0.5}}}~\cdot (1.20~m)$
$r_f = \sqrt{\frac{1}{\sqrt{10}}}~\cdot (1.20~m)$
$r_f = 0.675~m$
