The simplest nontrivial (unstable) integral cohomology operation

By an integral cohomology operation I mean a natural transformation $H^i(X, \mathbb{Z}) \times H^j(X, \mathbb{Z}) \times ... \to H^k(X, \mathbb{Z})$, where we restrict $X$ to some nice category of topological spaces such that integral cohomology $H^n(-, \mathbb{Z})$ is represented by the Eilenberg-MacLane spaces $K(\mathbb{Z}, n)$. The Yoneda lemma shows that such operations are in natural bijection with elements of $H^k(K(\mathbb{Z}, i) \times K(\mathbb{Z}, j) \times ... , \mathbb{Z})$.

Altogether, these cohomology operations determine a (multisorted) Lawvere theory. There is an obvious subtheory of this theory generated by zero, negation, addition, the cup product, and composition. Based on the table in this paper, it looks like the simplest integral cohomology operation not in this obvious subtheory is a cohomology operation $H^3(X, \mathbb{Z}) \to H^8(X, \mathbb{Z})$ coming from the generator of $H^8(K(\mathbb{Z}, 3), \mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z}$, but I don't know enough algebraic topology to extract an explicit description of this cohomology operation from the paper.

So: what is this cohomology operation? Where does it come from? What can you do with it? (I know there are some cohomology operations coming from the $\text{Tor}$ terms in the Künneth formula; is this one of them?)

Suppose $\alpha$ is an integral $3$-cocycle on a space $X$. Then $2\alpha\cup\alpha$ is a coboundary, because the cup product is graded-commutative, and there exists a $5$-cochain $\beta$ such that $\mathrm d\beta=2\alpha\cup\alpha$. One can easily check that $\alpha\cup\beta+\beta\cup\alpha$ is an $8$-cocycle. I think that the class of this cocycle does not depend on the choice of $\beta$, so we get a well defined mapping $H^3\to H^8$. This map is natural because there is a natural proof that the cup product is graded-commutative, so we $\beta$ depends naturally on $\alpha$.

This gives an operation :-)