I think the question is asking what the release altitude from a space elevator would result in an orbit with a perigee of $R_E + 500 km$ ($R_E$ is the earth radius — 6378.135 km)
The velocity of the space elevator varies linearly along its length, with the velocity at GEO altitude ($R_{GEO}$) equal to GEO orbital velocity ($V_{GEO} = \sqrt{\mu/R_g}$). So for a position $r$ on the elevator:
$$V_{elev} = {r\over R}\sqrt{\mu/R_{GEO}}$$
This velocity vector is perpendicular to the position vector, as is the orbital velocity vector at apogee (and perigee).
Orbital velocity as a function of radius $r$ is given by:
$$V_r = \sqrt{\mu(1/r-1/a)}$$
where $a$ is the semi-major axis
$$a = {R_{apogee} + R_{perigee}\over 2}$$
The release point will be the apogee of the achieved orbit, so we can re-write the equations as such:
$$V_{elev} = {R_{apogee}\over R_{GEO}}\sqrt{\mu/R_{GEO}}$$
$$V_{apogee} = \sqrt{\mu(1/R_{apogee} - 2/(R_{apogee} + R_{perigee}))}$$
We can now set $V_{elev} = V_{apogee}$
$${R_{apogee}\over R_{GEO}}\sqrt{\mu/R_{GEO}} = \sqrt{\mu(1/R_{apogee} - 2/(R_{apogee} + R_{perigee}))}$$
After some algebra, you can solve for $R_{perigee}$ as a function of $R_{apogee}$
$$R_{perigee} = (R_{apogee}^3/(2 R_{GEO}^3))[1/(1/R_{apogee} - R_{apogee}^2/(2 R_{GEO}^3))]$$
Solving for $R_{apogee}$ would be a bit messier I think, so I plugged this into Excel and used solver to numerically find the solution for $R_{perigee} = R_E + 500 = 6878.135$
$$R_{apogee} = 30276.54 km$$
Note that this is the radius from the earth center, not the height, as was used in the original question. Other quantities used:
$\mu = 398601$
$R_{GEO} = 42164.2 km$
