How to find the radius of a satellite not knowing the mass If I know the semi-major axis, the density of the planet, and orbital period.. how can I find the radius of the satellite?  I can't think of any formula to use.
How to find the radius of a satellite not knowing the mass:
If I know the

semi-major axis,
density of the planet, and
orbital period..

how can I find the radius of the satellite? I can't think of any formula to use.

I have a hunch that you've misunderstood the question, and that it asks for the radius of the planet not the artificial satellite, and that's certainly doable. Otherwise, why would the planet's density be a given?

The equation that links semi-major axis, mass of the planet, and orbital period is:
$$T=2 \pi \sqrt{\frac{a^3}{GM}} $$
where $G$ is the gravitational constant 6.674E-11 m^3 s^-2  kg^-1.
Moving things around we get
$$\frac{4 \pi^2 a^3}{T^2G} = M $$
Density is mass per volume, or:
$$\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}$$
Moving that around we get
$$R = \left(\frac{3M}{4 \rho \pi}\right)^{1/3}$$
Put in the earlier result for $M$, we get
$$R = a \left(\frac{3 \pi}{\rho T^2G}\right)^{1/3}$$

For Earth's average density, Wikipedia says 5514 kg/^3, for its radius, try 6,371,000. meters. For the ISS at 400 km (400,000 meters) above the Earth, use 92.5 minutes (5550 sec). It should all work out nicely!
The reason this works with average density, even though Earth's density is higher in the core than the surface is the application of Newton's shell theorem.