 Solution

If $R$ is reduced, its nilradical $N$ is zero. By 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 annihilatorChoose 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 $s\in {\mathrm{Ann}}_R(x)\subseteq \mathfrak{m}$, a contradiction. This is the principle that localization detects zero elements. ThereforeThe argument also includes the zero ring, which is reduced and has no maximal ideals.