jmc

$n^3\equiv 888(\mathrm{mod}1000)⟹n^3\equiv 0(\mathrm{mod}8)$ and $n^3\equiv 13(\mathrm{mod}125)$. $n\equiv 2(\mathrm{mod}5)$ due to the last digit of $n^3$. Let $n=5a+2$. By expanding, $125a^3+150a^2+60a+8\equiv 13(\mathrm{mod}125)⟹5a^2+12a\equiv 1(\mathrm{mod}25)$. By looking at the last digit again, we see $a\equiv 3(\mathrm{mod}5)$, so we let $a=5a_1+3$ where $a_1\in {\mathbb{Z}}^{+}$. Plugging this in to $5a^2+12a\equiv 1(\mathrm{mod}25)$ gives $10a_1+6\equiv 1(\mathrm{mod}25)$. Obviously, $a_1\equiv 2(\mathrm{mod}5)$, so we let $a_1=5a_2+2$ where $a_2$ can be any non-negative integer. Therefore, $n=2+5(3+5(2+5a_2))=125a_2+67$. $n^3$ must also be a multiple of $8$, so $n$ must be even. $125a_2+67\equiv 0(\mathrm{mod}2)⟹a_2\equiv 1(\mathrm{mod}2)$. Therefore, $a_2=2a_3+1$, where $a_3$ is any non-negative integer. The number $n$ has form $125(2a_3+1)+67=250a_3+192$. So the minimum $n=\boxed{192}$.