Answer
$420\space km/s$
Work Step by Step
Here we use the principle of conservation of mechanical energy.
Final mechanical energy = Initial mechanical energy
$K+U=K_{0}+U_{0}$
$-\frac{GM_{S}m}{r}+\frac{1}{2}mV^{2}=\frac{1}{2}mV_{0}^{2}-\frac{GM_{S}m}{R}$
Where m is mass of the material, r is the distance to the final position from the sun's center & $R$ is the distance to the initial position from the sun's center, V is the speed at the earth orbit, $V_{0}$ is the initial speed.
$V^{2}=V_{0}^{2}+2GM_{S}(\frac{1}{r}-\frac{1}{R})$
$V^{2}=(550\times10^{3}m/s)^{2}+2\times6.67\times10^{-11}Nm^{2}/kg^{2}\times1.99\times10^{30}(\frac{1}{150\times10^{9}m}-\frac{1}{3\times696\times10^{6}m})$
$V^{2}=3025\times10^{8}m^{2}/s^{2}+\frac{2.67\times10^{20}}{10^{6}}(6.7\times10^{-6}-479\times10^{-6})m^{2}/s^{2}$
$V^{2}=1764\times10^{8}m^{2}/s^{2}$
$V=42\times10^{4}m/s=420\space km/s$
