= Solution
If $R$ is <reduced>, its <nilradical> $N$ is zero. By <3/b>, every <localization at a prime ideal> has zero <nilradical>, so in particular $R_{\mathfrak m}$ is <reduced> for every <maximal ideal> $\mathfrak m$.
Conversely, suppose every such <localization> is <reduced>. Take a <nilpotent element> $x\in R$. Then $x/1$ is <nilpotent> in every $R_{\mathfrak m}$, hence zero there. To see directly that $x=0$, suppose otherwise and consider the proper <annihilator>
$$
\operatorname{Ann}_R(x)=\{a\in R:ax=0\}.
$$
Choose a <maximal ideal> $\mathfrak m$ containing this <annihilator>. The equality $x/1=0$ in $R_{\mathfrak m}$ would imply $sx=0$ for some $s\notin\mathfrak m$, by the <vanishing criterion in a module localization>. But then $s\in\operatorname{Ann}_R(x)\subseteq\mathfrak m$, a contradiction. This is the principle that <localization detects zero elements>. Therefore
$$
\boxed{R\text{ is reduced}\iff R_{\mathfrak m}\text{ is reduced for every maximal ideal }\mathfrak m.}
$$
The argument also includes the <zero ring>, which is <reduced> and has no <maximal ideals>.
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