$(a)\;0.707\;w/m^2\;,2.52\times10^{5} \;m\qquad(b)\;0.707\;w/m^2\;,2.52\times10^{3} \;m\qquad(c)\;0.707\;w/m^2\;,2.52\times10^{4} \;m\qquad(d)\;None$

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Answer : (c) $\;0.707\;w/m^2\;,2.52\times10^{4} \;m$

Explanation :

$I=\large\frac{P av}{4 \pi r^2}=\large\frac{80.0w}{4 \pi (3.00m)^2}=0.707w/m^2$

$10 log(\large\frac{I}{I_{0}})=40 dB$

$log I-log I_{0}=\large\frac{40}{10}=4$

$I_{0}=1.00\times 10^{-12} w/m^2$

$log I=4 + log 10^{-12}$

$log I=-8$

$I=1.00 \times 10^{-8} w/m^2$

Using this value for I in equation r = $\; \sqrt{\large\frac{Pav}{4 \pi I}}$

$=\sqrt{\large\frac{80.0 w}{4 \pi \times 1.00 \times 10^{-8} w/m^2}}$

$=2.52 \times 10^{4} m$

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