Correct Answer - C
`dB = 10 log_(10) [(I)/(I_0)]`,
where `I_0 = 10^(-12) wm^-2`
Since, `40 = 10 log_(10) [(I_1)/(I_0)]`
`rArr (I_1)/(I_0) = 10^4`
Also, `20 = 10 log_(10) [(I_2)/(I_0)]`
`rArr (I_2)/(I_0) = 10^2`
So, `(I_2)/(I_1) = 10^-2 = (r_1^2)/(r_2^2)`
`rArr r_2^2 = 100 r_1^2`
`rArr r_2 = 10 m`.