Answer
(a) $6MW/m^2$
(b) less than
(c) $1MW/m^2$
Work Step by Step
(a) The required intensity can be determined as
$I_{av}==\frac{mg\times c}{\pi r^2}$
We plug in the known values to obtain:
$I_{av}=\frac{(1.6\times 10^{-15}Kg)(9.8m/s^2)(3\times 10^8m/s)}{\pi(0.5\times 10^{-6}m)^2}$
$I_{av}=5.988\times 10^6W/m^2=6MW/m^2$
(b) If the radius is doubled then $r^{\prime}=2r$
Now $I^{\prime}_{av}=\frac{mg\times c}{\pi (r^{\prime})^2}$
$I^{\prime}_{av}=\frac{mg\times c}{4(\pi r^2)}$
$I^{\prime}_{av}=\frac{1}{4}I_{av}$
Thus, the intensity is less than the value found in part(a).
(c) From part (b), we have
$I^{\prime}_{av}=\frac{1}{4}I_{av}$
$\implies I^{\prime}_{av}=\frac{1}{4}(5.988\times 10^{6})=1.4\times 10^6W/m^2=1MW/m^2$
