 Solution

Taking the Earth's sidereal rotation period as $86164\mathrm{s}$, $\omega =2\pi /(86164\mathrm{s})$, and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ givesThe orbital radius is therefore about $6.61R$, or an altitude of about $5.61R$. Replacing the sidereal day by $86400\mathrm{s}$ changes this estimate by less than $0.2\%$.