The rate of decrease of mass $m$ of a black hole due to Hawking radiation goes as
$$\frac{dm}{dt} = -\frac{\hbar c^4}{15360 \pi G^2} \frac{1}{m^2} \,, $$
where the symbols have their usual meaning. If you assume that the mass growth is totally due to (isotropic) accretion of baryonic matter or cold-dark matter, then the mass growth takes the form (see Bondi-Hoyle accretion)
$$\frac{dm}{dt} = \frac{4 \pi G^2 m^2 \rho_{\infty}} {\left(c_{\infty}^2+v_{\infty}^2\right)^{3 / 2}} \,,$$
where $\rho_\infty$ is the density of the medium a large distance from the black hole, $v_\infty$ is the speed at which the black hole movies w.r.t. the static medium and $c_\infty$ is the speed of the sound in the medium.
A black hole would be stable against Hawking evaporation if the second equation dominates the first one. To find the answer you were looking, you should equate the above two equations. This will give
$$m = \left[ \frac{\hbar c^4}{15360} \frac{\left(c_{\infty}^2+v_{\infty}^2\right)^{3 / 2}} {4 \pi^2 G^4 \rho_{\infty}} \right]^{1/4}$$
You can assume $v_{\infty} = 0$ but would need to know the other values of the ambient medium.
For the medium you asked about, air at STP, the values are:
$c_{\infty} = 331 m/s$, and $\rho_{\infty} = 1.29 kg/m^{3}$.
Using these numbers results in a black hole with the following mass and Schwarzschild radius: $$m = 3.8×10^{10} kg$$ $$r_{s} = 5.6×10^{-17} m$$
However, as noted in the comments, this Schwarzschild radius is too small for the gas to be treated as a perfect fluid. A more accurate value of the mass limit in this medium would require an equation other than the standard Bondi-Hoyle equation, which accounts for the quantum-mechanical nature of gas particles at that scale.
