Answer
The ratio of the larger power to the smaller power is $~~3.16$
Work Step by Step
We can consider the sound levels at $r = 1000~m$
Let $I_A$ be the greater sound intensity.
We can find $I_A$:
$\beta = (10~dB)~log(\frac{I_A}{I_0})$
$70.0~dB = (10~dB)~log(\frac{I_A}{I_0})$
$7.00 = log(\frac{I_A}{I_0})$
$10^{7.00} = \frac{I_A}{I_0}$
$I_A = I_0 \cdot~10^{7.00}$
$I_A = (10^{-12}~W/m^2)~(10^{7.00})$
$I_A = 10^{-5.0}~W/m^2$
Let $I_B$ be the smaller sound intensity.
We can find $I_B$:
$\beta = (10~dB)~log(\frac{I_B}{I_0})$
$65.0~dB = (10~dB)~log(\frac{I_B}{I_0})$
$6.50 = log(\frac{I_B}{I_0})$
$10^{6.50} = \frac{I_B}{I_0}$
$I_B = I_0 \cdot~10^{6.50}$
$I_B = (10^{-12}~W/m^2)~(10^{6.50})$
$I_B = 10^{-5.5}~W/m^2$
We can find the ratio of the larger power to the smaller power $\frac{P_A}{P_B}$:
$\frac{P_A}{P_B} = \frac{I_A~4\pi r^2}{I_B~4\pi r^2}$
$\frac{P_A}{P_B} = \frac{I_A}{I_B}$
$\frac{P_A}{P_B} = \frac{10^{-5.0}~W/m^2}{10^{-5.5}~W/m^2}$
$\frac{P_A}{P_B} = 10^{0.5}$
$\frac{P_A}{P_B} = 3.16$
The ratio of the larger power to the smaller power is $~~3.16$
