The period of revolution of an earth satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at an altitude of three times the radius of earth is

Distance of second satellite from the center of earth becomes four times
Since, according to Kepler’s Laws we have ${\text{T}}^2\propto \mathrm{\hspace{0.17em}}{\mathrm{r}}^3\Rightarrow \mathrm{T}\propto {\mathrm{r}}^{3/2}$.
So, time period of second satellite must become ${\left(4\right)}^{3/2}=8$ times.
Hence $\mathrm{T}\text{'}=8\mathrm{T}=\left(8\right)\left(90\right)=720\min$.