What are the missing digits?

Let $x$ be the answer to this question. From the question, we know that $x^3\equiv888\mod{1000}\Rightarrow x\equiv2\mod{10}$

Now, let $x=10a+2$ Substituting back into the original equation, we have $(10a+2)^3\equiv888\mod{1000}\\1000a^3+600a^2+60a+8\equiv888\mod{1000}\\600a^2+120a\equiv 880\mod{1000}$

This means that $60a^2+12a$ ends in 8. Trying all possible remainders of $a$ modulo 10, we find that only 9 satisfies this equivalence. Therefore, let $a=10b+9\Rightarrow x=100b+92$ . Then, we have $(100b+92)^3\equiv888\mod{1000}\\1000000b^3+2760000b^2+2539200b+778688\equiv888\mod{1000}\\200b+688\equiv888\mod{1000}$

Clearly, the smallest possible value of $b$ is 1, so the answer is 192.