Solution

The power radiated by the Sun is $P$, so the intensity of the radiation at a distance $r$ from the Sun is:

\begin{equation} I = \frac{P}{4 \pi r^2} \end{equation}

Thus, the pressure exerted by the light on the dust particle is:

\begin{equation} \mathcal{P} = 2 \frac{I}{c} = \frac{P}{4 \pi r^2 c} \end{equation}

The corresponding force is:

\begin{equation} F = 2 \pi r_1^2 \mathcal{P} = \frac{P r_1^2}{2 r^2 c} \end{equation}

This force must be at least equal to the gravitational force:

\begin{equation} F = \frac{G M_s m_d}{r^2} \quad \text{where} \quad m_d = \rho \frac{4 \pi r_1^3}{3} \end{equation}

Finally:

\begin{equation} \frac{G M_s m_d}{r^2} = \frac{P r_1^2}{4 r^2 c} \rightarrow r_1 = \frac{3 P}{8 \pi M_s G c \rho} \approx 1 \, \mu\text{m} \end{equation}