= Solution
Let $N$ be the <nilradical> of $R$. By <3/b>, the <reduced locus of a ring> is
$$
\mathcal R=\{\mathfrak p\in\operatorname{Spec}R:NR_{\mathfrak p}=0\}.
$$
The <Noetherian ring> hypothesis makes the <ideal> $N$ finitely generated; write $N=(n_1,\ldots,n_t)$. Fix $\mathfrak p\in\mathcal R$. For every $i$, the <vanishing criterion in a module localization> gives $s_i\notin\mathfrak p$ with $s_in_i=0$. The product $f=s_1\cdots s_t$ lies outside the <prime ideal> $\mathfrak p$ and satisfies $fN=0$. When $N=0$, take $f=1$.
For every <prime ideal> $\mathfrak q$ in the <distinguished open set>
$$
D(f)=\{\mathfrak q\in\operatorname{Spec}R:f\notin\mathfrak q\},
$$
the element $f/1$ is a <unit> in $R_{\mathfrak q}$. The equality $fN=0$ therefore implies $NR_{\mathfrak q}=0$, so $R_{\mathfrak q}$ is <reduced>. Thus $\mathfrak p\in D(f)\subseteq\mathcal R$, proving that $\mathcal R$ is a <Zariski-open subset> of the <spectrum of a ring>.
More precisely, let $A=\operatorname{Ann}_R(N)$, the <annihilator of a module> $N$. The argument shows that $NR_{\mathfrak p}=0$ if and only if some $f\in A$ lies outside $\mathfrak p$. Hence
$$
\boxed{\mathcal R=\bigcup_{f\in\operatorname{Ann}_R(N)}D(f)
=\operatorname{Spec}R\setminus V(\operatorname{Ann}_R(N)).}
$$
Equivalently, the complement is the <support of a module> $N$. This is <vanishing of a finitely generated module near a prime> applied to $N$: its finite generation lets the separately chosen denominators be combined into one element annihilating all of $N$.
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