If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the center of the planet will become

The correct answer is 2⁰.⁶⁶⁷ times the original distance.

Let's understand the relation between the orbital radius of a geostationary satellite and the angular velocity of the planet.
A geostationary satellite orbits the planet with an orbital period equal to the rotation period of the planet about its own axis. Thus, the angular velocity of the satellite's orbit ($\omega$) is equal to the angular velocity of the planet's rotation.

For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center of the planet, the gravitational force provides the necessary centripetal force:
$\frac{GMm}{r^2}=m{\omega }^{2}r$

Simplifying this equation for the orbital radius $r$:
$r^3=\frac{GM}{{\omega }^2}$
This gives:
$r\propto {\omega }^{-2/3}$

Let the initial angular velocity be ${\omega }_1$ and the initial distance be $r_1$.
The new angular velocity is halved:
${\omega }_2=\frac{{\omega }_1}{2}$

Using the proportionality relation, the new distance $r_2$ is given by:
$\frac{r_2}{r_1}$
Substituting ${\omega }_2=\frac{{\omega }_1}{2}$:
$\frac{r_2}{r_1}$

Since $\frac{2}{3}\approx 0.667$, we have:
$r_2=2^{0.667}{r}_1$

Therefore, the new distance of the geostationary satellite from the center of the planet will become 2⁰.⁶⁶⁷ times the original distance.