c
(c) \(dB = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\); where \({I_0} = {10^{ - 12}}W{m^{ - 2}}\)

Since \(40 = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right)\)==> \(\frac{{{I_1}}}{{{I_0}}} = {10^4}\) ....\((i) \)

Also \(20 = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right)\) ==>\(\frac{{{I_2}}}{{{I_0}}} = {10^2}\) ....\((ii) \)

==> \(\frac{{{I_2}}}{{{I_1}}} = {10^{ - 2}} = \frac{{r_1^2}}{{r_2^2}}\)

==> \(r_2^2 = 100r_1^2\)==> \({r_2} = 10m\)          \(\{\because {r_1} = 1m\} \)