= Solution
For small $\theta$,
$$
1-\cos\theta=\frac{\theta^2}{2}\left(1-\frac{\theta^2}{12}+O(\theta^4)\right),
\qquad
\theta-\sin\theta=\frac{\theta^3}{6}\left(1-\frac{\theta^2}{20}+O(\theta^4)\right).
$$
Put $u=(6t/B)^{1/3}$. Inverting the second series gives
$$
\theta=u\left(1+\frac{u^2}{60}+O(u^4)\right).
$$
Substitution into the first series yields
$$
R=\frac{AR_0}{2}\left(\frac{6t}{B}\right)^{2/3}
\left[1-\frac1{20}\left(\frac{6t}{B}\right)^{2/3}+O(t^{4/3})\right].
$$
Using the definitions of $A$ and $B$ simplifies its leading factor to
$$
\frac A2\left(\frac6B\right)^{2/3}t^{2/3}
=\Omega_{m,0}^{1/3}\left(\frac{3H_0t}{2}\right)^{2/3}
=\Omega_{m,0}^{1/3}a(t).
$$
Therefore
$$
\boxed{R(t)=R_0\Omega_{m,0}^{1/3}a(t)
\left[1-\frac1{20}\left(\frac{6t}{B}\right)^{2/3}+O(t^{4/3})\right].}
$$
The homogeneous sphere containing the same mass has radius $\bar R=R_0\Omega_{m,0}^{1/3}a(t)$. By mass conservation, its nonlinear <density contrast> is
$$
1+\delta=\left(\frac{\bar R}{R}\right)^3
=1+\frac3{20}\left(\frac{6t}{B}\right)^{2/3}+O(t^{4/3}).
$$
The leading, <linear cosmological density perturbation> is consequently
$$
\boxed{\delta_{\rm linear}(t)=\frac3{20}\left(\frac{6t}{B}\right)^{2/3}\propto t^{2/3}\propto a(t).}
$$
In the parametric solution, complete spherical collapse occurs at $\theta=2\pi$, when $R=0$ and $t=B(2\pi-\sin2\pi)=2\pi B$. Extrapolating the linear growing mode to that time gives the standard <linear spherical-collapse threshold>
$$
\boxed{\delta_{\rm linear,coll}
=\frac3{20}(12\pi)^{2/3}\simeq1.686.}
$$
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