= Solution
For small <baryon> fraction the CDM growing mode is $\delta_C\propto a$, and its gravity forces
$$
\delta_B''+\mathcal H\delta_B'+c_s^2k^2\delta_B\simeq4\pi Ga^2\bar\rho_C\delta_C.
$$
For an <adiabatic cosmological perturbation> to follow the unsuppressed growing mode, <pressure> must be below the gravitational forcing frequency: $c_s^2k^2\lesssim4\pi Ga^2\bar\rho_C$. With physical wavelength $\lambda=2\pi a/k$, this becomes
$$
\boxed{\lambda\gtrsim\lambda_J=c_s\sqrt{\frac\pi{G\bar\rho_C}}}.
$$
This is <Jeans support versus forced baryon response>. The wording “only grow” must refer to unsuppressed gravitational structure growth; it is not a theorem that all shorter-wavelength <baryon> perturbations are absent or constant. In the pressure-dominated regime a forced response $\delta_B\simeq(k_J/k)^2\delta_C$ remains. In particular, a falling <sound speed> can make this small response grow even before it catches up to CDM.
Before <photon decoupling>, the <photon-baryon fluid> has <sound speed> of order $c/\sqrt3$. Its Jeans scale is roughly a horizon scale, so smaller perturbations oscillate acoustically instead of freely following the growing CDM structures. After decoupling the <baryon> <sound speed> drops sharply. Above the new Jeans scale, <pressure> is negligible and subtraction of the CDM equation gives $(\delta_B-\delta_C)''+\mathcal H(\delta_B-\delta_C)'=0$. For $a\propto\tau^2$, the difference is a constant plus a decaying $1/\tau$ term; its ratio to the growing CDM contrast tends to zero. <Baryons> therefore catch up. Under the stipulated post-decoupling $c_s\propto T\propto a^{-1}$, the physical <Jeans length> scales as $a^{1/2}$, the comoving scale as $a^{-1/2}$ and the corresponding <Jeans mass> as $a^{-3/2}$.
For the requested instantaneous <mass> estimates, define the baryonic <Jeans mass> as the <baryon> content of a sphere of radius $\lambda_J/2$:
$$
M_{JB}=\frac\pi6\bar\rho_B\lambda_J^3,\qquad\bar\rho_B\simeq0.1\bar\rho_C.
$$
Using the supplied matter-era age gives $t_{\rm dec}=t_0(1+z_{\rm dec})^{-3/2}\simeq9.5\times10^{13}$ s and $\bar\rho_C\simeq1/(6\pi Gt_{\rm dec}^2)\simeq6\times10^{-23}\ {\rm g\,cm^{-3}}$. Then
$$
\boxed{M_{JB}^{\rm before}\sim6\times10^{18}M_\odot\quad(c_s=c/\sqrt3),\qquad M_{JB}^{\rm after}\sim3\times10^4M_\odot\quad(c_s=10^{-5}c)}.
$$
The drop is by $(\sqrt3\times10^{-5})^3$. A definition using the total gravitating <mass> instead of <baryon> content gives values about ten times larger. The pre-decoupling number is an acoustic Jeans order estimate, not an accurate application of the nonrelativistic equations from part a when $c_s\sim c$.
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