A signal of 0 . 1kW is transmitted in a cable. The attenuation of...

Sound level decreases by $5dB$ every $km$ so, sound level decreased in $20km = 100dB$

${\beta}_{2} - {\beta}_{1} = 10\log_{10}\dfrac{{I}_{2}}{{I}_{1}}$

$- 100 = 10\log_{10}\dfrac{{I}_{2}}{{I}_{1}} ⇒ \dfrac{{I}_{1}}{{I}_{2}} = {10}^{10}$

${I}_{2} = {10}^{- 10}{I}_{1} ⇒ {P}_{2} = {10}^{- 10}{P}_{1} = {10}^{- 8}W$

$x = 8$