jmc

Let $m^3$ be a positive five-digit perfect cube with an 8 in the ten-thousands place. Thus, $8\times 10^4<m^3<9\times 10^4⟺20\sqrt[3]{10}<m<10\sqrt[3]{90}\Rightarrow 43<m<45$. Therefore we conclude $m=44$, so $m^3=44^3=\boxed{85,184}$.