Answer
$P = 4.68\times 10^{11}~W$
Work Step by Step
We can find the mass of the sphere:
$M = V~\rho$
$M = \frac{4}{3}\pi~R^3~\rho$
$M = \frac{4}{3}\pi~(2.00\times 10^{-3}~m)^3~(19,000~kg/m^3)$
$M = 6.367\times 10^{-4}~kg$
We can find the required power of the light source:
$F = Mg$
$\frac{I~A}{c} = Mg$
$\frac{P~\pi~R^2}{4\pi~r^2~c} = Mg$
$P = \frac{4Mg~r^2~c}{R^2}$
$P = \frac{(4)(6.367\times 10^{-4}~kg)(9.8~m/s^2)~(0.500~m)^2~(3.0\times 10^8~m/s)}{(2.00\times 10^{-3}~m)^2}$
$P = 4.68\times 10^{11}~W$
