Given a mass (energy) density of $\rho$, a ball of radius $R$ has a mass of $M(R)=4\pi/3\cdot R^3\cdot\rho$. The Schwarzschild Radius of this mass in dependence of $R$ is $$ r_s(R) = \frac{2GM(R)}{c^2} = \frac{2G\cdot 4\pi\cdot R^3\cdot\rho}{c^2\cdot 3} = \frac{8\pi G R^3\rho}{3 c^2} $$
Now we can ask, given a fixed $\rho$, when does the Schwarzschild radius get larger than $R$, i.e. when do we have $r_s(R)\ge R$? That would be if \begin{align} \frac{8\pi G R^3\rho}{3 c^2} &\ge R\\ R^2 &\ge \frac{3c^2}{8\pi G\rho} \end{align} For the average density of the universe Wikipedia lists $\rho=9.9\times 10^{-27} \text{kg}/\text{m}^3$. We also have $G=6.7\times 10^{-11} \text{m}^2/(\text{s}^2\text{kg})$ and $c=299792458\, \text{m}/\text{s}$. Inserting the values we get that a volume with radius $R$ and the mass density of the universe passes its Schwarzschild radius if
\begin{align} R^2 &\ge \frac{3\cdot 299792458^2}{8\pi \cdot 6.7\times 10^{-11} \cdot 9.9\times 10^{-27}} \text{m}^2\\ R^2 &\ge 1.62\times 10^{52} \text{m}^2 \\ R &\ge 1.27 \times 10^{26} \text{m}\\ \text{or in light years}\\ R &\ge 13.44\times 10^9 \text{Ly} \end{align}
Question 1: Where did I screw up above? This seems to contradict this answer whereby the Schwarzschild Radius of the whole mass of the universe is around $500\times 10^9 \text{Ly}$? (I did this in three different ways and always got the avobe result. :-/)
Edit Question 2: So far nobody pointed out formal errors in the above, so I dare to ask: how do we explain away that any sufficiently large sub-ball of the universe has a radius smaller than its Schwarzschild Radius?
