The derivation and background here isn't really important. The reader should skip to the bottom of this post for the actual question before reading the background.
Background
In Modern Cosmology by Dodelson and Schmidt 2nd edition, we are trying to find an equation for the Compton scattering rate $n_{e}\sigma_{T}$ divided by the Hubble rate $H$. Equation (4.43) reads $$ \frac{n_{e}\sigma_{T}}{H} = 123X_{e}\left(\frac{\Omega_{\text{b}}h^{2}}{0.022}\right)\left(\frac{0.14}{\Omega_{\text{m}}h^{2}}\right)^{1/2}\left(\frac{1+z}{1000}\right)^{3/2}\left[1 + \frac{1+z}{3360}\frac{0.14}{\Omega_{\text{m}}h^{2}}\right]^{-1/2}. $$
To derive this, let us take equation (4.42) for granted as a starting point: $$ \frac{n_{e}\sigma_{T}}{H} = 0.0692 \,a^{-3}X_{e}\Omega_{\text{b}}h\frac{H_{0}}{H}. $$
By the first Friedmann equation, we have $$ \frac{H}{H_{0}} = \frac{\left(\frac{8\pi G}{3} \rho\right)^{1/2}}{H_{0}} = \left( \frac{8\pi G}{3H_{0}^{2}} \rho \right)^{1/2} = \left( \frac{\rho}{\rho_{\text{cr}}} \right)^{1/2}. $$
Now assuming radiation and matter only, we have
\begin{align*} \frac{\rho}{\rho_{\text{cr}}} &= \frac{\rho_{\text{r}} + \rho_{\text{m}}}{\rho_{\text{cr}}} \\ &= \Omega_{\text{r}}a^{-4} + \Omega_{\text{m}}a^{-3} \\ &= \Omega_{\text{m}} a^{-3} [1 + \frac{\Omega_{\text{r}}}{\Omega_{\text{m}}}a^{-1}] \\ &= \Omega_{\text{m}} a^{-3} [1 + a_{\text{eq}}/a] \end{align*} where $a_{\text{eq}}$ is the scale factor at which the density of radiation matches the density of matter. This gives $$ \frac{H}{H_{0}} = \Omega_{\text{m}}^{1/2}a^{-3/2}[1 + a_{\text{eq}}/a]^{1/2}. $$
Now we put this back into (4.42) all the way above, and we have the following:
\begin{align*} \frac{n_{e} \sigma_{T}}{H} &= 0.0692 \,a^{-3} X_{e} \Omega_{\text{b}} h \frac{H_{0}}{H} \\[1.6ex] &= 0.0692 X_{e} \cdot a^{-3}\Omega_{\text{b}}h \cdot \Omega_{\text{m}}^{-1/2}a^{3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.0692 X_{e} \cdot \Omega_{\text{b}} h \cdot \Omega_{m}^{-1/2} \cdot a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.0692 X_{e} \cdot \Omega_{\text{b}} h^{2} \cdot\left(\Omega_{\text{m}} h^{2}\right)^{-1/2} a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.00407 X_{e} \cdot\left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} a^{-3/2}\left[1+a_{\text{eq}} / a\right]^{-1/2} \\[1.6ex] &= 0.00407 X_{e} \left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} \left(1+z\right)^{3/2}\left[1+(1+z) a_{\text{eq}}\right]^{-1/2} \\[1.6ex] &= 128.7 X_{e}\left(\frac{\Omega_{b} h^{2}}{0.022}\right) \cdot\left(\frac{0.14}{\Omega_{m} h^{2}}\right)^{1/2} \left(\frac{1+z}{1000}\right)^{3/2} \left[1+\frac{1+z}{3360} \cdot 3360 a_{\text{eq}}\right]^{-1/2}. \end{align*}
The prefactor number $128.7$ seems to be off from the $123$ prefactor in the book, but that doesn't concern me. In order to match the equation in the book, I need to take $$ 3360a_{\text{eq}} = \frac{0.14}{\Omega_{\text{m}}h^{2}}. $$
Question
Is there a reason why $$ 3360a_{\text{eq}} = \frac{0.14}{\Omega_{\text{m}}h^{2}}? $$ How is this derived? What info do you need to obtain this? Is there some typo or is there a significance to the number $3360$?