This is exercise 12.3 O from Vakil's FOAG, self-study.
We are to show that if $X$ is a regular (locally Noetherian) scheme (meaning its local rings are all regular local rings), then so is $X \times \mathbb A^1$.
The idea is to use the previous proposition 12.3.6, which says that if $(B, \mathfrak n)$ is a regular local ring and $\phi: B \to A := B[x]$, then for any prime $P \subset A$ such that $\mathfrak n A \subset P$, $A_P$ is a regular local ring.
I interpreted $X \times \mathbb A^1$ to mean $X \times_{\mathbb Z} \mathbb A^1_{\mathbb Z}$, and assumed $X = \operatorname{Spec} B$, where $B$ is Noetherian regular local. Then our fibered product gives us a map
$$B \to B \otimes \mathbb Z[x] \simeq B[x]$$
By the proposition, if we localize $B[x]$ at any primes which are big enough to contain the extension of $\mathfrak n$, we obtain a regular local ring, so our fibered product is regular only at those points. I am unsure how to conclude it is regular at all points. We only know smoothness can be checked at closed points, but not regularity.
Best Answer
Your interpretation of $X\times\Bbb A^1$ to mean $X\times_{\Bbb Z} \Bbb A^1_{\Bbb Z}$ is correct. The reduction to the affine case is good, but you don't have to assume that $B$ is local, and I think this is actually maybe leading you astray here.
What do we know? We know that for every prime $\mathfrak{p}\subset B$, the ring $B_\mathfrak{p}$ is a regular local ring. What do we want to prove? That for every prime ideal $\mathfrak{q}\subset B[x]$, the ring $(B[x])_\mathfrak{q}$ is a regular local ring. Since $\mathfrak{q}\cap B$ is a prime ideal of $B$, we can apply the proposition to $B_{\mathfrak{q}\cap B} \to (B[x])_{\mathfrak{q}}$ as you've done and get that $(B[x])_\mathfrak{q}$ is regular, and we're done.