Why every smooth orientable 4-dimensional manifold admits an immersion into $\mathbb{R}^{6}$?

Why every smooth orientable 4-dimensional manifold admits an immersion into $\mathbb{R}^{6}$?

This is a one-line question as I see the statement in a comment by Michael Hopkins(update: this is wrong, it is by Peter Kronheimer) . I thought about it for a long time but I do not know how to prove it or to approach it. Characteristic classes provide a way of showing "if this...", but does not help to show the existence of such an immersion. (comment: this is stupid line of thinking because obviously I did not make use of all characteristic tools available to me, as evident in reading Kirby's book)

update: The correct statement is provided in Kirby's book, page 44 Lemma 1, which states such immersion exists iff there exists a characteristic class $x\in H^{2}(M^{4};\mathbb{Z})$ such that $x_{(2)}=-w_{2}$ and $x^{2}=-p_{1}$.

I suspect you either mis-read the Hopkins statement, or Hopkins made an error. Likely the statement he intended is that every orientable 4-manifold is cobordant to one that immerses in $\mathbb R^6$ -- this is in Kirby's book on 4-manifolds and is a commonly used step in the proof of Rochlin's theorem.

But not every orientable 4-manifold immerses in $\mathbb R^6$, as $w_4$ is an obstruction. This is a theorem of K. Sakuma's. Sakuma Reference

So for example, $\mathbb CP^2$ does not immerse in $\mathbb R^6$ according to Sakuma, since it has odd Euler characteristic.

This is the main topic of Chapter VI (entitled "Immersing $4$-manifolds in $\mathbb{R}^6$") of Kirby's book "The Topology of 4-Manifolds".

I emailed Michael Hopkins and he said this problem set is not his, but composed by Peter Kronheimer. So the content of my question is quite inappropriate and I apologize for the misnomber in here. I will leave the question unchanged (for otherwise the answers may be incomprehensible). I will update this "answer" once I worked out Kirby's proof.