1. Use the geostationary condition

For a geostationary satellite, the orbital time period equals the rotation period of the planet: $T_1 = 24\text{ h}$

It is given that this satellite is at a height of $11R$ above the surface.

So its distance from the center of the planet is: $r_1 = R + 11R = 12R$

2. Find the radius of the second satellite's orbit

The second satellite is at a height of $2R$ above the surface, so: $r_2 = R + 2R = 3R$

3. Apply Kepler's third law for satellites around the same planet

For satellites orbiting the same planet, $T^2 \propto r^3$

Hence, $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$

Substitute the values: $\left(\frac{T_2}{24}\right)^2 = \left(\frac{3R}{12R}\right)^3$ $\left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$

Taking square root, $\frac{T_2}{24} = \frac{1}{8}$ $T_2 = 24 \times \frac{1}{8} = 3\text{ h}$

4. Match with the options

Thus, the time period of the second satellite is: $\boxed{3\text{ h}}$

So the correct option is A.