The time period of a satellite in a circular orbit of the radius $$R$$ is $$T$$. The period of another satellite in a circular orbit of the radius $$9R$$ is:

By Kepler's third law, the square of the orbital period of a satellite is proportional to the cube of the orbital radius: $$T^2 \propto R^3$$. This can be written as $$T = k R^{3/2}$$ for some constant $$k$$ that depends on the central body's mass.

For the first satellite with orbital radius $$R$$ and period $$T_1 = T$$, and the second satellite with orbital radius $$R_2 = 9R$$ and period $$T_2$$, we can write $$\frac{T_2}{T_1} = \left(\frac{R_2}{R_1}\right)^{3/2} = \left(\frac{9R}{R}\right)^{3/2} = 9^{3/2}$$.

Now $$9^{3/2} = (9^1)(9^{1/2}) = 9 \times 3 = 27$$. Alternatively, $$9^{3/2} = (3^2)^{3/2} = 3^3 = 27$$.

Therefore $$T_2 = 27T_1 = 27T$$. The period of the satellite at radius $$9R$$ is $$27T$$.