Firstly we should note that the universe as a whole is not described by the Schwarzschild metric, so the Schwarzschild radius of the universe is a meaningless concept. However if you take the mass of the observable universe you could ask what the Schwarzschild radius of a black hole of this mass is.
For a mass $M$ the Schwarzschild radius is:
$$ r_s = \frac{2GM}{c^2} \tag{1} $$
If the radius of the observable universe is $R$, and the density is $\rho$, then the mass is:
$$ M = \tfrac{4}{3}\pi R^3 \rho $$
and we can substitute in equation (1) to get:
$$ r_s = \frac{8G}{3c^2} \pi R^3 \rho \tag{2} $$
Now we believe that the density of the universe is the critical density, and from the FLRW metric with some hair pulling we can obtain a value for the critical density:
$$ \rho_c = \frac{3H^2}{8\pi G} $$
And we can substitute for $\rho$ in equation (2) to get:
$$ r_s = \frac{H^2}{c^2} R^3 \tag{3} $$
Now, Hubble's law tells that the velocity of a distant object is related to its distance $r$ by:
$$ v \approx Hr $$
and since the edge of the universe, $r_e$, is where the recession velocity is $c$ we get:
$$ r_e \approx \frac{c}{H} $$
and substituting this in equation (3) gives;
$$ r_s = \frac{1}{r_e^2} R^3 \tag{4} $$
If $r_e = R$ then we'd be left with $r_s = R$ and we'd have shown that the Schwarzschild radius of the mass of the observable universe is equal to it's radius. Sadly it doesn't quite work. The dimension $R$ is the current size of the observable universe, which is around 46.6 billion light years, while the size used in Hubble's law, $r_e$, is the current apparent size 13.7 billion light years.
If I take equation (3) and put in $R$ = 46.6 billion light years and $H$ = 68 km/sec/megaParsec I get $r_s$ to be around 500 billion light years or a lot larger than the size of the observable universe.