Answer
$7.07\times10^{5}\;W/m^2$
Work Step by Step
The average intensity in the incident beam
$I=\frac{P}{A}=\frac{P}{\pi d^2/4}$
Substituting the given values, we obtain
$I=\frac{5\times4}{\pi \times (3\times 10^{-3})^2}\;W/m^2=7.07\times10^{5}\;W/m^2$
Therefore, The average intensity in the incident beam is $7.07\times10^{5}\;W/m^2$
