This is a followup to my previous question, reproduced here.
Let$$0 \to R \to F \to G \to 0$$be a short exact sequence of groups. Is it possible to construct an associated fibration of spaces$$K(R, 1) \to K(F, 1) \to K(G, 1)?$$
Qiaochu Yuan gave the following answer.
Yes. In fact a fiber sequence of Eilenberg-MacLane spaces is the same thing as a short exact sequence of groups.
Take a surjective group homomorphism $f: G \to H$. This induces a map $Bf : BG \to BH$ on classifying spaces/Eilenberg-MacLane spaces (how to see this at the point-set level depends on what your point-set construction of Eilenberg-MacLane spaces is). Now take the homotopy fiber of $Bf$. A computation with the long exact sequence in homotopy shows that it is $B\text{ker}(f)$.
My question is, how do we calculate the $E_{p, 0}^2$ and the $E_{0, 1}^2$ terms in this sequence, in terms of the homology of $K(G, 1)$, the homology of $K(R, 1)$, and the action of $G$ on $R$ by conjugation?