Answer
$10^{1.4}:1$
Work Step by Step
Loudness of a sound with intensity $I, L=10log_{10}(\frac{I}{I_{0}})$
Consider intensity of the rock music =$ I$
And intensity of the mower = $ I'$
As per the given problem, we have
$10log_{10}(\frac{I}{I_{0}})=120$ dB
$log_{10}(\frac{I}{I_{0}})=12$ dB
or
$I=(10)^{12}{I_{0}}$ ...(1)
Amplified rock music is measured at 120 dB, whereas the noise from a motor-driven lawn mower is measured at 106 dB.
$10log_{10}(\frac{I'}{I_{0}})=106$ dB
$log_{10}(\frac{I'}{I_{0}})=10.6$ dB
or
$I=(10)^{10.6}{I_{0}}$ ...(2)
From equations 1 and 2, we get
$\frac{I}{I'}=\frac{(10)^{12}}{(10)^{10.6}}$
This implies,
$I:I'=10^{1.4}:1$
Hence, the ratio of the intensity of the rock music to that of the mower is $I:I'=10^{1.4}:1$.
