Question

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a period of exactly 24.0 hours

Answer 1

So they've given you the period of the circular motion of the satellite.
From $$\omega=\frac{2\pi}T$$,
you can find the angular velocity of the satellite.
now, for a circular motion to happen,the centrifugal force(F) should be,
$$F=mr\omega^2$$
Now this F should be generated from the gravitational field on the satellite. That gravitational force is,
$$F=\frac{GmM}{r^2}$$
where M is the mass of the earth.
equating these two forces, we get
$$\begin {align*} mr\omega^2&=\frac{GmM}{r^2}\\
r\omega^2&=\frac{GM}{r^2}\\
r&=\sqrt[3]{\frac{GM}{\omega^2}}\\
r&=\sqrt[3]{\frac{GMT^2}{4\pi^2}} \end {align*}$$