$M_s=2 \times 10^{30} \mathrm{~kg} . \quad M_e=6 \times 10^{24} \mathrm{~kg} r=1.5 \times 10^{11} \mathrm{~m}$ Let at a distance $x$ from the earth the gravitational force on the rocket due to sun and the earth are equal and opposite.
$\therefore$ Distance of the rocket from the sun $=r-x$
If $m$ is the mass of the rocket then
$$
\begin{aligned}
& \frac{G M_s m}{(r-x)^2}=\frac{G M_e m}{x^2} \Rightarrow \frac{(r-x)^2}{x^2}=\frac{M_s}{M_e} \\
\Rightarrow & \frac{r-x}{x}=\sqrt{\frac{M_s}{M_e}} \Rightarrow \sqrt{\frac{2 \times 10^{30}}{6 \times 10^{24}}}=\frac{10^3}{\sqrt{3}} \\
\Rightarrow & \frac{r}{x}-1=\frac{10^3}{\sqrt{3}} \Rightarrow \frac{r}{x}=\frac{10^3}{\sqrt{3}}+1=\frac{10^3+\sqrt{3}}{\sqrt{3}} \\
\therefore \quad & x=\frac{\sqrt{3} r}{10^3+\sqrt{3}}=\frac{1.732 \times 10^{11} \times 1.5}{10^3+1.732} \\
&=2.59 \times 10^8 \mathrm{~m}
\end{aligned}
$$