= Solution
A <nilpotent element> belongs to every <prime ideal> and therefore to every <maximal ideal>, so
$$
\sqrt{(0)}\subseteq\operatorname{Jac}(R).
$$
For the reverse containment, let $r$ be nonnilpotent. The <localization of a ring> $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a <finite-type integer algebra>, because it can be presented as $R[T]/(rT-1)$.
Choose a <maximal ideal> $\mathfrak n$ of $S$. Its <residue field> $L=S/\mathfrak n$ is a <finite-type integer algebra>, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite <integral domain> containing $1$, so it is a <field>: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.
The kernel $\mathfrak m$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero <residue field>. Thus $r\notin\mathfrak m$. Every nonnilpotent element can therefore be avoided by a <maximal ideal>, giving
$$
\boxed{\operatorname{Jac}(R)=\sqrt{(0)}}.
$$
This <radical equality for finitely generated integer algebras> uses part (a) to ensure that the contraction of the localized <maximal ideal> is maximal. Such contraction is not generally maximal for arbitrary <localization of a ring>; the finite-field image is the decisive additional step. The zero <ring> is immediate, with the intersection of its empty set of <maximal ideals> understood as the whole <ring>. No general theorem that integer algebras are <Jacobson rings> is being quoted.
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