You are asking a wrong question. Here is the problem with your reasoning.
You are assuming a Schwarzschild metric and a homogenous distribution of mass. But the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter. For a cosmological spacetime filled with matter, like our universe, the suitable metric to use would be something else, like the FRW for example.
You could only use the Schwarzschild spacetime if you assumed a sphere of some uniform density $\rho$ and vacuum outside the radius of the sphere.
Let me illustrate how things would work out then. As you can see, a particular density corresponds to a particular $R_s$, lets call it $R_s(\rho)$. So if you had a sphere of matter with a radius $R_1$ grater than $R_s(\rho)$, then you couldn't apply the formula $R_s(\rho)=c\sqrt{\frac{3}{8\pi G \rho}}$. You would have to use the Schwarzschild metric only in the vacuum region outside of the sphere. So you would have then $R_s=\frac{8\pi G\rho R_1^3}{3c^2}$. In order to see how the $R_s$ compares with $R_s(\rho)$, you can replace the density with $\rho=\frac{3c^2}{8\pi G R_s(\rho)}$. So you would get that the Schwarzschild radius for a sphere of uniform density $\rho$ and radius $R_1>R_s(\rho)$ is $R_s=\left(\frac{R_1}{R_s(\rho)}\right)^2R_1$, which is grater than the radius of the sphere. So the sphere is inside its Schwarzschild horizon. If on the other hand, the radius $R_1$ is smaller than $R_s(\rho)$, then the corresponding horizon would have to be inside the sphere. But inside the sphere the Schwarzschild metric doesn't apply. So it isn't necessary that there should be a horizon inside the matter distribution.
If you apply these to the universe and assume for example that the radius of the visible universe is the radius $R_1$ of the sphere, then you would have a horizon radius (using your numbers) that would be almost 10 times the radius of the observable universe. So, the entire universe would have to be in a black hole of radius of 460 billion light-years. So the assumption that we should see black holes with horizons of radii of 13.9 billion light-years is not correct.
If one assumes the above point of view, one could say that the universe is a white hole that is exploding.
I hope that all these are helpful and not confusing.
The basic problem is that for a sufficiently massive star, the electrons become relativistic. The fine details of this calculation are rather complicated, but you can get a qualitative sense of the argument as follows:
For non-relativistic fermions at zero temperature, it is possible to show that the total energy of $N$ particles in a box of volume $V$ is proportional to $N^{5/3}/V^{2/3}$. This can be done via counting the density of states, and using the fact that the energy of a non-relativistic particle obeys $E \propto |\vec{p}|^{2}$.
For a spherical volume of radius $R$, we have $R \propto V^{1/3}$, and the number of fermions present is proportional to the mass. This means that the total energy of the fermions is proportional to $M^{5/3}/R^2$. This energy is positive.
On the other hand, the gravitational energy of a solid sphere is negative and proportional to $M^2/R$. This means that the total energy is the sum of a negative $R^{-1}$ term and a positive $R^{-2}$ term, and such a function will have a minimum somewhere. This will be the equilibrium point. At smaller radii, the energy of the degeneracy grows faster than the binding energy decreases, pushing the radius back to larger values. At larger radii, the reverse occurs. This means that the star will be stable.
This argument doesn't hold up to arbitrarily large energies, though, because eventually the Fermi energy of the electrons exceeds the rest energy of the electron; in other words, the electrons become relativistic. This changes the relationship between energy and momentum of the electrons. For highly relativistic electrons, we have $E \propto |\vec{p}|$ instead; and going through the same calculations (neglecting the electron mass entirely), we find that the total energy of a relativistic fermion gas is proportional to $N^{4/3}/V^{1/3} \propto M^{4/3}/R$.
The gravitational binding energy, on the other hand, remains negative and proportional to $M^2/R$. This implies that the overall energy is itself proportional to $1/R$, and there is no extremum of the total energy of the system. Since the fermion energy and the binding energy always increase or decrease at exactly the same rate, there will be no stable equilibrium radius. The star will either blow itself apart or collapse in on itself, depending on whether the kinetic energy of the fermions or the gravitational binding energy wins out.
Best Answer
Primordial black holes were never stars at any stage in their lifecycle, so those mass limits do not apply to them. See this Wikipedia article.