For a hefty, substantial spacecraft like the 5000 kg, 5 square meter, drag coefficient = 1 simulation I did here, the final complete orbit was circularized at about 95 kilometers. A cube sat of the same average density would have been higher just because of the higher area to mass ratio.
For the atmosphere to match other tabulated values around 100km I used the very very rough approximation
$$\rho(h) \ = \ \rho_0 \ exp(-h/h_{scale})$$
With the sea-level density and scale height $\rho_0$ and $h_{scale}$ of 1.225 kg/m^3 and 7700 meters. This is just rough.
A simple equation for drag from Wikipedia might be
$$F_{drag} \ = \ -\frac{1}{2} \ \rho \ C_D \ A \ v^2$$
and $C_D$ for a randomly shaped satellite at LEO velocity might be 1, the area might be 5 meters, and the LEO velocity $v$ might be 7700 m/s.
$$F_{drag} \ = \ 0.5 \times 1.225 \times exp(-h/h_{scale}) \times 1.0 \times 5.0 \times 7700^2$$
If I assume the thruster has a force of 0.2 N as I did here, I can solve for the height $h$, which in this case turns out to be almost exactly your original number, 146 km.
Did I say roughly, and say it enough times?
Now the problem is that the upper atmosphere changes a lot in response to activity of the Sun, and so the density is not well approximated by a simple scale height, or is even constant or predictable. This is discussed in @DavidHammon's answer here and comment here. So if there is a solar event, you may suddenly need maybe even ten times the thrust to keep from burning up all of a sudden. So while estimates can give you a feel for it, your mileage may vary, a lot!
People talk about putting spy or Earth monitoring satellites at altitudes between 150 and 200 km because a smaller telescope (like a 10cm aperture inside a 3U cubesat) might get better resolution than one at 400km, or similar to a 50cm aperture at 750 km. Of course there are a lot of other factors that affect resolution, and there is a problem with atomic oxygen eating your satellite, but people still talk about it.
