- What is the explicit realization of following Eilenberg-MacLane spaces? There are some examples I know but not for the higher
$${\displaystyle K(\mathbb {Z} ,1)}=S^1$$ $${\displaystyle K(\mathbb {Z} ,2)}=\mathbb {CP}^\infty$$ $${\displaystyle K(\mathbb {Z} ,3)}=B\mathbb {CP}^\infty=?$$
$${\displaystyle K(\mathbb {Z} /2,1)}=\mathbb {RP}^\infty.$$ $${\displaystyle K(\mathbb {Z} /2,2)}=B\mathbb {RP}^\infty=?$$ $${\displaystyle K(\mathbb {Z} /2,3)}=B^2\mathbb {RP}^\infty=?$$
- Why does it mean to state there a weak equivalence ${\displaystyle TOP/PL\sim K(\mathbb {Z} /2,3)}$ of TOP/PL with an Eilenberg–MacLane space?
For your first question, there are many reasonable constructions of these. One way is via linearization of spheres (see the section at https://ncatlab.org/nlab/show/Eilenberg-Mac+Lane+space). Another way to think about it is through the Dold-Kan correspondence. In this lens, an Eilenberg-MacLane space is just the space associated to the chain complex with that group concentrated in that degree.
I can say more about the second question:
Recall that $Top$ is the colimit of the $Top(n)$ which is the homeomorphism group of $\mathbb{R}^n$. Similarly, $PL$ is the colimit of the $PL(n)$ which is the PL homeomorphism group of $\mathbb{R}^n$. I will first go over the argument of Kirby and Siebenmann, and then I will explain how it relates to Wikipedia's statement.
Now usually it is the case that when we prove a stabilization result, it is easier to get at the stabilized and then use the result to get at the unstable object. Here Kirby and Siebenmann go the other way around. They prove the result that the inclusion $Top(m)/PL(m) \rightarrow Top/PL$ is an isomorphism on the first $m$ homotopy groups if $m>4$ and will use this to get at the homotopy groups of the stable object. So with this result is suffices to prove that actually if $m>4$ that $Top(m)/PL(M)$ is a $K(\mathbb{Z}/2,3)$.
This is done by using a very great result of Morlet that $PL_\partial(D^m) \simeq \Omega^{m+1}(Top_m/PL_m)$. The space on the left is contractible by what is called the Alexander trick, so we get that the homotopy groups above $m$ are trivial. This result holds for $m \neq 4$. So if we take $m=5$ and manually check that the homotopy groups below $6$ are what we want, then we are done. This is their theorem 5.3 and is rather long. My understanding is that they compare triangulations (see the latter half of my answer) and "homotopy triangulations". The latter are accessible through surgery type arguments.
So using these three results: stability, triviality of high homotopy groups, and having the correct lower homotopy groups, we see that $Top/PL$ is a $K(\mathbb{Z}/2,3)$. Now how does this relate to what wikipedia says?
Well the reason we care about $Top$ and $PL$ so much is that their classifying spaces classify stable Top and PL $\mathbb{R}^n$ bundles. These are like vector bundles, but there is no vector space structure on the fibers. From the fiber sequence $Top/PL \rightarrow BPL \rightarrow BTop$ we can deduce that the homotopy groups of $Top/PL$ are important in asking when we can lift a Top bundle to a PL bundle and to what extent this is unique.
But in fact these spaces are even more important than just for these reasons. The question of whether a manifold can be given a PL structure, and if this is unique, is actually equivalent (if we are not in dimension 4) to whether or not the stable tangent bundle of the manifold (this is not a vector bundle if we are not smooth) can be given a PL structure. This might be called the fundamental theorem of triangulation theory.
By obstruction theory, we know that the obstructions to lifting a map along a fibration lie in the cohomology of the domain with coefficients in the homotopy groups of the fiber, shifted by 1. In our case we are using the fibration $Top/PL \rightarrow BPL \rightarrow BTop$, if it were the case that every homotopy group of $Top/PL$ was realized as an obstruction to lifting a map $[M, BTop]$ to a map $[M,BPL]$ (I am not sure how one could argue this without Kirby and Seibenmann's result), for some topological manifold M, then the fact that the Kirby-Siebanmann obstruction is the only obstruction and lies in $H^4 (M;\mathbb{Z}/2)$ would imply that $Top/PL$ is a $K(\mathbb{Z}/2,3)$. But to be clear, this is not how the result was proven, and I believe the Kirby-Siebanmann invariant is defined with the knowledge that the fiber is an Eilenberg-MacLane space.