A satellite is called stationary or synchronous if its period (i.e. the time required for
one orbital revolution) is one day. What is the height above the Earth’s surface at which a synchronous satellite needs to be placed?
Kepler's 3rd law of planetary motion (derived from Newton's Law of Universal Gravitation = centripetal force) ...
$\displaystyle T^2 = \dfrac{4\pi^2}{GM_e} \cdot R^3
$
$\displaystyle R = R_e + h$
you need to solve for the orbital radius, $\displaystyle R$ , then determine the value of $\displaystyle h$
A similar reasoning: The centripetal acceleration $\displaystyle v^2/(R+h)=\omega^2(R+h)$ is provided by the gravity, so it equals $\displaystyle GM/(R+h)^2$. Here R is the Earth's radius, M is the Earth's mass, h is the satellite's height above the Earth’s surface, v is the satellite's velocity, $\displaystyle \omega$ is the satellite's angular velocity and G is the gravitational constant. The angular velocity is given.