Solution

Let $I$ be the intensity of light from the Sun on the Earth and the satellite. Consequently, the power of light from the Sun intercepted by the cross section of the satellite is $I\cdot\pi r^2$, where $r$ is the radius of the satellite. Since the light is reflected by a half of the satellite, the intensity of light from the satellite on the Earth is

$$\frac{I\cdot\pi r^2}{2\pi d^2}=I\cdot\frac{1}{2}\left(\frac{r}{d}\right)^2,$$

where $d$ is the distance between the satellite and the Earth. Using $r=1$ m and $d=300$ km, we see that the intensity of the light on the Earth from the satellite is $1/(18\times10^{10})$ of that from the Sun. Since light intensity varies inversely with the square of the distance from the source, a star comparable to the Sun will give the same intensity of light on the Earth as the satellite if it is $\sqrt{18\times10^{10}}\approx4.2\times10^5$ times as far from the Earth as the Sun. Because the Sun is about $500$ light-second from the Earth, the star is about $500\times4.2\times10^5$ light-seconds, which is roughly $6.6$ light-years, from the Earth. Note that the Sun as a yellow dwarf is in fact brighter than most stars which are red dwarfs, so a better estimate of the distance is less than $6.6$ light-year.