Boundary of product manifolds such as $S^2 \times \mathbb R$

As others have written, $S^2\times\mathbb R$ has no boundary.

Generally, if $A$ and $B$ are manifolds with boundaries $\partial A$ and $\partial B$, respectively, then $A\times B$ is a manifold with boundary $\partial(A\times B)=(\partial A\times B)\cup(A\times\partial B)$. This is true for manifolds but not for differential manifolds, since the "corners" are homeomorphic but not diffeomorphic to $\mathbb R^{n-1}$. For instance, the product of two closed line segments is a closed rectangle, which is homeomorphic but not diffeomorphic to a closed disc. See this paper for an exposition on manifolds with corners.

If you are interested in differentiable manifolds, a variation of what joriki said is still true. If $X$ is a manifold without boundary and $Y$, a manifold with boundary $\partial Y$, then $X\times Y$ is a manifold with boundary $X\times \partial Y$. For more details, you can look at many places, e.g. Guillemin and Pollack Differential Topology, Chapter 2.

It has no boundary. Let $p=(\theta,\psi,r)\in M:=S^2\times\mathbb{R}$ (think spherical coordinates) then $p$ is an interior point. Indeed $p$ is contained in the open set (a "cube"): $$(\theta-\epsilon,\theta+\epsilon)\times(\psi-\epsilon,\psi+\epsilon)\times(r-\epsilon,r+\epsilon)$$

As an illustrative aside, note also that you can represent M visually as $\mathbb{R}^3-0$. Indeed, if you just associate $r\in\mathbb{R}$ with a sphere of radius $e^r$, then you get the desired visual of $\mathbb{R}^3$ minus the origin. (Note that as $r$ gets bigger so does the radius, and as $r$ gets more negative the radius shrinks.)