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

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

Edit: Even if a black hole is slightly lighter than the above mass, it would still take an extremely long time to evaporate completely (on the order of $10^{40}$ years). And if the black hole is heavier than the above mass, it will still evaporate, but only after the CMB cools down sufficiently ($10^{100}$ years for supermassive black holes). With these kinds of time scales, the notion of "stability" starts to blur a bit!

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:

"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

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