The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$
Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$
If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.
The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$
For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes
$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields
$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$ Unfortunately, this does not seem to allow us to obtain an actual number for our estimate, since $H_\text{em}$ is not known...
