cosmological inflation - What was the age of the universe when the average density was one atmosphere?

The critical density of the universe is
$$rho_c=frac{3H_0^2}{8pi G},$$
with $H_0=67.8 frac{mbox{km}/mbox{s}}{mbox{Mpc}}$ the Hubble constant and $G=6.673848cdot 10^{-11}frac{mbox{m}^3}{mbox{kg }mbox{s}^2}$ Newton's gravitational constant.
Hence with one parsec $1mbox{pc}=3.0857×10^{16} mbox{m}$,
$$rho_c=frac{3cdot (67.8cdot 10^3cdot frac{mbox{m}/mbox{s}}{3.0857×10^{22} mbox{m}})^2}{8picdot 6.673848cdot 10^{-11}frac{mbox{m}^3}{mbox{kg}mbox{s}^2}}=8.635 cdot 10^{-27}frac{mbox{kg}}{mbox{m}^3}.$$

Ordinary matter content is $4.9%$, the universe has total density close to critical density, hence ordinary matter density is
$$0.049cdot 8.635 cdot 10^{-27}frac{mbox{kg}}{mbox{m}^3}=4.23 cdot 10^{-28}frac{mbox{kg}}{mbox{m}^3}.$$
(The Hubble constant is not precisely known, so the calculated density depends a bit on the precise value.)

Density of water:
The density of water is about $10^3 mbox{kg}/mbox{m}^3$.
That's a factor $4.23 cdot 10^{31}$ denser than the average universe.
The space needs to have been reduced by a factor $sqrt[3]{4.23 cdot 10^{31}}=3.48cdot 10^{10}$ in each if the three spatial dimensions. That's the inverse of the scale factor, hence corresponds to a redshift of $z=3.48cdot 10^{10}$.

Using the Cosmology Calculator on this website, the cosmological parameters $H_0 = 67.11$ km/s/Mpc (slightly different to the above value),
$Omega_{Lambda} = 0.6825$ provided by the Planck project, and setting $Omega_M = 1- Omega_{Lambda} = 0.3175$ the age of the universe was 0.022 seconds at the redshift $z=3.48cdot 10^{10}$, hence when it was of the average density of water.

Density of air at sea level:
The density of air at sea level is $1.225 frac{mbox{kg}}{mbox{m}^3}$. The corresponding redshift is about $z=1.423cdot 10^9$ yielding an age of the universe of 12.68 seconds.