= Solution
Suppose, for a contradiction, that $X=M\times\mathbb R$ has a complete metric with zero <Ricci curvature>. Since $M$ is closed and <connected>, the complement of $M\times[-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are <compact> by the <Hopf-Rinow theorem>. Thus $X$ is <disconnected at infinity>, and part (a) supplies a line. The <Cheeger-Gromoll splitting theorem> gives an <isometry>
$$
X\cong N^3\times\mathbb R.
$$
The <Riemannian product> curvature formula makes $N$ complete and Ricci-flat. A three-dimensional <Ricci-flat Riemannian manifold> is flat: for any orthonormal triple, the three trace equations are $\operatorname{Ric}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}=\operatorname{Ric}_{ii}+\operatorname{Ric}_{jj}-\operatorname{Scal}/2=0$. Every plane can occur in such a triple, so every <sectional curvature> vanishes, and hence so does the <Riemann curvature tensor>.
The complete <simply connected> <universal cover> of $N$ is therefore Euclidean $\mathbb R^3$, and that of $X$ is Euclidean $\mathbb R^4$. But the topological <universal cover> of $X$ is $\widetilde M\times\mathbb R$. Its <deformation retraction> onto $\widetilde M\times\{0\}$ shows that contractibility of this cover would imply contractibility of $\widetilde M$, contrary to the hypothesis. Therefore \b[no complete Ricci-flat metric exists on $M\times\mathbb R$]. Notice that no product assumption was made about the hypothetical original metric.
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