 Solution

Set $S=A\setminus \mathfrak{m}$, a multiplicative subset because $\mathfrak{m}$ is a prime ideal. The quotient module $B=A/\mathfrak{a}$ is a cyclic module generated by $\bar{1}$, so its localization of a module $S^{-1}B$ is zero precisely when $\bar{1}/1=0$. By the vanishing criterion in a module localization, this is equivalent to the existence of $s\in S$ with $s\bar{1}=0$ in $B$. The latter condition says $s\in \mathfrak{a}$. ThereforeExplicitly, if $s\in \mathfrak{a}\setminus \mathfrak{m}$, multiplication by $s$ annihilates $B$ but becomes invertible after localization, forcing every localized element to be zero.