Answer
${\bf 0.585}\;\rm \mu m$
Work Step by Step
We need to find the radius of a dust particle that can remain in the solar system. This dust particle will remain when the gravitational force exerted on it is greater, or at least equal to, the radiation force.
$$F_G=F_{\rm radiation}$$
$$\dfrac{Gm_{D}m_S}{r^2}=PA\tag 1$$
where $m_D$ is the dust mass, $m_S$ is the sun mass, $r$ is the distance between the sun and the particle, $P$ is the radiation pressure, and $A=\pi R_D^2$ is the cross-sectional area of the dust particle.
The mass of the dust particle is given by the density law
$$m_D=\rho V=\frac{4}{3}\rho_D \pi R_D^3\tag 2$$
We know that the radiation pressure on an object that absorbs the light is given by
$$P=\dfrac{I}{c} $$
where $P$ is the pressure, and $I={\rm Power}/A$ is the intensity of the radiation.
The particle is black, so it absorbs all the radiation that falls on it.
$$P=\dfrac{{\rm Power}}{4\pi cr^2} \tag 3$$
Plug (2) and (3) into (1),
$$\dfrac{\frac{4}{3}G\rho_D \pi R_D^{ \color{red}{\bf\not} 3}m_S}{ \color{red}{\bf\not} r^2}= \dfrac{{\rm Power}}{ 4\color{red}{\bf\not} \pi c \color{red}{\bf\not} r^2} \cdot \color{red}{\bf\not} \pi \color{red}{\bf\not} R_D^2 $$
Hence, the radius of the dust is given by
$$R_D=\dfrac{{\rm Power}}{ \frac{16\pi}{3}G\rho_D cm_S}$$
where its diameter is given by $D_D=2R_D$,
$$D_D=\dfrac{2\;{\rm Power}}{ \frac{16\pi}{3}G\rho_D cm_S}$$
Plug the known;
$$D_D=\dfrac{2(3.9\times 10^{26})}{ \frac{16\pi }{3}(6.67\times 10^{-11}) (3\times 10^8)(2000) (1.99\times 10^{30})}$$
$$D_D=\color{red}{\bf 5.845\times 10^{-7}}\;\rm m$$
This is the minimum diameter of particles that can remain for a long time in our solar system.
