 Solution

The geostationary gravitational clock gain uses only the potential contribution, as requested. In the weak field, $d\tau /dt\simeq 1+\Phi /c^2$ and $\Phi =-GM/r$. The geostationary radius follows from $GM/r_g^3={\Omega }_{\oplus }^2$. Taking $GM\simeq g{R}^2$ with $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $R=6.4\times 10^6\mathrm{m}$ and sidereal day $86164\mathrm{s}$ gives $r_g\simeq 4.23\times 10^7\mathrm{m}$.The satellite clock gains time because it is at higher gravitational potential. This is not the combined gravitational-plus-kinematic correction; the orbital and surface rotational velocities are deliberately outside the requested potential contribution.