Answer
$R = 5.8\times 10^{-7}~m$
Work Step by Step
Let $r$ be the radius of the particle.
To find the critical radius, we can equate the magnitude of the gravitational force and the force from the radiation pressure:
$\frac{G~M_s~M}{d^2} = \frac{I~A}{c}$
$\frac{G~M_s~V~\rho}{d^2} = \frac{I~\pi~r^2}{c}$
$\frac{G~M_s~\frac{4}{3}\pi~r^3~\rho}{d^2} = \frac{I~\pi~r^2}{c}$
$\frac{4~G~M_s~r~\rho}{3d^2} = \frac{I}{c}$
The critical radius $R$ is the value of $r$ which makes this equation true.
We can find $R$:
$\frac{4~G~M_s~R~\rho}{3d^2} = \frac{I}{c}$
$\frac{4~G~M_s~R~\rho}{3d^2} = \frac{P}{4~\pi~d^2c}$
$R = \frac{3P}{16~\pi~c~G~M_s~\rho}$
$R = \frac{(3)(3.90\times 10^{26}~W)}{(16~\pi)~(3.0\times 10^8~m/s)~(6.67\times 10^{-11}~N~m^2/kg^2)~(2.0\times 10^{30}~kg)~(1000~kg/m^3)}$
$R = 5.8\times 10^{-7}~m$
