If $M$ is a hyperfinite type III factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$.
This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.
• The group of units of $M$ is contractible.
• The automorphism group of $M$ is contractible (conjectural).
To see that $Out(M)\cong (\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: $$ U(Z(M)) \to U(M) \to Inn(M) $$ $$ Inn(M) \to Aut(M) \to Out(M) $$
As a consequence, we also get that $BOut(M)\cong K(\mathbb Z,4)$.