In order to be "stable", the black hole's Hawking radiation temperature would need to be equal to the temperature of the cosmic microwave background, which is currently 2.7 K.

From [Wikipedia][1]:
>"A black hole of $4.5 × 10^{22}$ kg (about the mass of the Moon) would be in equilibrium at 2.7 kelvin, absorbing as much radiation as it emits"

So then, the Schwarzschild radius of such a black hole would be:

$r_\mathrm{s} = \frac{2G(4.5 × 10^{22})}{c^2}$

= 0.00007m


  [1]: http://en.wikipedia.org/wiki/Hawking_radiation