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Dmitry Brant
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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. (Assuming this is what you meant by "stable"?)

From Wikipedia:

"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

Dmitry Brant
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