A smooth (i.e. $C^{\infty}$) $n$-manifold $M$ can be defined as a topological manifold such that each point has a neighborhood which is diffeomorhic to an open subset of $\mathbb{R}^n$. In particular, every point has a neighborhood which is diffeomorphic to $\mathbb{R}^n$ (see e.g. here).
This definition is unambiguous if $n\neq 4$, since then $\mathbb{R}^n$ has a unique smooth structure. However, in dimension 4, Euclidean space $\mathbb{R}^4$ has uncountably many incompatible smooth structures.
Suppose $M$ is a smooth 4-manifold and $p$ is a point in $M$. Then $p$ has a neighborhood $U$ which is diffeomorphic to $\mathbb{R}^4$, but which smooth $\mathbb{R}^4$? Is the convention that the smooth structure on $\mathbb{R}^4$ is taken to be the standard one?
Are there smooth 4-manifolds such that any point has a neighborhood diffeomorphic to an exotic $\mathbb{R}^4$ (other than the exotic $\mathbb{R}^4$s themselves)? In other words, can one construct exotic 4-manifolds from exotic $\mathbb{R}^4$s? (For example, if you take a quotient $\mathbb{R}^4/\mathbb{Z}^4$ of an exotic $\mathbb{R}^4$, do you get an exotic torus?)
It is always the standard $\mathbb{R}^4$.
Also notice that the non-standard $\mathbb{R}^4$ are locally diffeomorphic and globally homeomorphic to the standard $\mathbb{R}^4$, but not globally diffeomorphic.