Today‘s scale factor of the universe How large is the non-normalized scale factor (solution of the Friedmann equation) in today‘s universe? How it is measured?
 A: The scale factor $a$ has an arbitrary normalization. Typically people take $a=1$ for the present time.
A: Because we don't know whether the Universe is finite or infinite, it does not make sense to talk about an absolute size, or radius, of the Universe. Instead we use an arbitrary, dimensionless scale called the scale factor, usually denoted $a$.
As the Universe expands, all (cosmological) distance scale with $a$. We define this to be equal to unity today, and say that, for instance, when the Universe was half the (linear) size, $a$ was equal to $0.5$. This number can then be multiplied on physical scales to give physical distance.
In principle you could use a non-normalized scale factor, i.e. with dimensions of length. But you would be completely free to choose whichever distance you wish, since an infinite universe doesn't have a preferred scale.
If the cosmological principle — which states that the Universe is homogeneous and isotropic on sufficiently large scales — holds true, then there are three possible geometries for the Universe: flat, open, and closed. Only the Universe is closed is it finite. In this case you could use a non-normalized scale factor equal to the radius of the Universe, which is equal the its radius of curvature. This is usually called $R_0$.
If the Universe were open, it's infinite, but still has an associated radius of curvature; however it would be an imaginary number, so it would make less sense to use as a standard for measuring distance (though you could normalize it).
As I discuss in this answer on astronomy.SE, a recent paper by Di Valentino et al. (2019) claims to have evidence (in the Planck 2018 data) for a closed Universe, with a curvature density parameter of roughly $\Omega_K\simeq-0.04$. This would correspond to a radius of the Universe of $R_0 \sim 67_{-21}^{+100}\,\mathrm{Glyr}$, not much larger than the observable Universe.
But the important thing is that, even if the Universe is finite and we can measure its radius, this would not make $R_0$ any more "correct" as the scale factor as any other distance you can come up with. So there really is no use for a non-normalized scale factor.
Moreover, $R_0$ will always have an associated error, whereas $a$, by definition, is errorless.
A: 
How large is the non-normalized scale factor (solution of the Friedmann equation) in today‘s universe? 

Scale factor by itself does not mean anything. The important thing is the ratios of the two scale factors. 
If we are describing the expansion of the universe in terms of the scale factor, the important question becomes, scale factor relative to what ?

How it is measured?

You cannot measure a scale factor. Its not a measureable quantity. What matters is that the ratios of the scale factor.
There's an equation that relates $z$ and $a(t)$,
$$1+z = \frac{a(t_0)} {a(t_e)}$$ 
Let us say we looked at an object and observed $z=3$ this means that 
$$4 = \frac{a(t_0)} {a(t_e)}$$
Here you can see that
$$\frac{a(t_0)} {a(t_e)} = \frac{4} {1} = \frac{10^{10}} {25 \times 10^8} = \frac{0.20} {0.05} = \frac{1} {0.25}$$
For $a(t_0)$ you can choose any value. Since its not an observable quantity it does not matter what you choose. But just for simplicity we choose $a(t_0) = 1$.
