1. Let the satellite orbit at a distance $r$ from the center of the Earth.

   If $h$ is the height from Earth’s surface, then $r=R+h$ where $R$ is Earth’s radius.

2. For a satellite in circular orbit, gravitational force provides the centripetal force: $\frac{GMm}{r^2}=m\omega^2 r$

   Using $\omega=\frac{2\pi}{T}$, $\frac{GM}{r^2}=\left(\frac{2\pi}{T}\right)^2 r$

3. Rearranging, $GM=\frac{4\pi^2 r^3}{T^2}$ so $r^3=\frac{GMT^2}{4\pi^2}$

4. Now use the relation at Earth’s surface: $g=\frac{GM}{R^2}$ hence $GM=gR^2$

   Substitute into the expression for $r^3$: $r^3=\frac{gR^2T^2}{4\pi^2}$

5. Therefore, $r=\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3}$

6. Since height above Earth’s surface is $h=r-R$, $h=\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3}-R$

7. Compare with the options:

   • A: $\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$ ✅
   • B: wrong power and wrong denominator
   • C: wrong sign of power and wrong final operation
   • D: $g$ is in denominator, which is incorrect

Therefore, the correct option is A.