A body in a stationary orbit will always remain above the same point on the planet as it orbits. For a body to be in such an orbit, it must rotate around the planet in the same direction as the spin of the planet, and its orbital period must be equal to the period of rotation of the planet. In this question we are aksed to calculate the orbital radius at which the satellite will complete one orbital cycle in precisely the time that it will take the planet to complete one full revolution.
The satellite, mass m, will be undergoing uniform circular motion around the centre of mass of the planet at some radius r. The centripetal force, FC, required to keep the satellite moving at a constant angular speed w (where w=2*pi/T; T is the orbital period of the satellite), will be given by
FC=mrw2.
But what gives rise to this centripetal force? Recall that centripetal force is not a force in itself, but rather the name for a force which always acts centrally (towards the same point) on a body undergoing circular motion. In this case, the central force is the gravitational pull of the planet on the satellitle, FG, such that FC=FG. By Newton's universal law of gravitation,
FG=(GMm)/r2,
where M is the mass of the planet, and G is the gravitational constant 6.67*10-11 m3kg-1s-2.
Equating the two forces together, we get that
mrw2 = (GMm)/r2.
We wish to find r, so rearranging to make r the subject and noticing that the mass of the satellite cancels out, we get that
r3 = (G*M)/w2.
We know that w=(2*pi)/T, and we also know that T must be equal to the period of rotation of the planet for a stationary orbit, which we are given. Making this substitution for w, and performing a little algebra,
r3 = (GMT2)/4*pi2.
If we substitute in the values of G, M and T, and take the cubed root to get r, we get that
r = 1.3*108 m.
Thus, for our satellite to be in a stationary orbit around this planet it must be 1.3*108 m away from the centre of the planet.