Answer
$2.49\times10^{10}\;W/m^2$
Work Step by Step
The central disk contains $84\%$ of the incident power.
The power of incident beam is $5\;W$
Therefore, the power contained in the central disk is
$P_c=5\times\frac{84}{100}\;W=4.2\;W$
The radius of the central disk is $7.33\times 10^{-6}\;m$
The average intensity in the central disk
$I=\frac{P_c}{A}=\frac{P}{\pi r_c^2}$
Substituting the given values, we obtain
$I=\frac{4.2}{\pi \times (7.33\times 10^{-6})^2}\;W/m^2=2.49\times10^{10}\;W/m^2$
Therefore, The average intensity in in the central disk is $2.49\times10^{10}\;W/m^2$
