This is not a complete answer as I won't be including the exact calculation needed to find out your burn time, but at least I will address the direct return vs bi elliptic approach.
For a return from orbit of a manned spacecraft, you want to balance two factors:
On one side you want to minimise the amount of fuel required for the operation; on the other, you want to minimise the time spent and the final speed.
The quickest and safest return would be a direct return using a Hohmann transfer orbit; burn retrograde at apogee and you will arrive to your destination orbit faster than using any other method and with the lowest reentry speed hence maximising your chance of survival.
If you use a Bi-elliptic transfer orbit you can reduce the total Delta V needed, but at the expense of more time in space for your astronauts and a higher reentry speed.
I've done a quick calculation, and a direct Hohmann transfer from a circular geostationary orbit to 100Km (and let the atmosphere do the rest) would require around 1.49Km/s (please someone confirm) and take 17 hours. A bi-elliptic going up to 380,000Km (Moon's distance to Earth, just to pick a meaningful distance for easy reference) would save you ~167m/s (11%) at the cost of 10 days in space.
Note after HopDavid's comment: Usually for this two orbits a bi-eliptic transfer should be less efficient, but as we are using the atmosphere for our final "burn", we save us the most expensive of them. The higher you go in the bi-eliptic, the more energy you have to shed in the final circularization burn, and the most fuel you save with aerobraking.