jmc

We have $$\begin{array}{rl}(a+bi{)}^3-107i& =(a^3+3a^{2}bi-3a{b}^2-b^{3}i)-107i\\ & =(a^3-3a{b}^2)+(3a^{2}b-b^3-107)i.\end{array}$$If this is a real number, then we must have $$0=3a^{2}b-b^3-107$$or $$107=b(3a^2-b^2).$$Since $107$ is prime, either $b=1$ or $b=107.$ If $b=1,$ then we have $107=3a^2-1,$ so $a^2=36$ and $a=6.$ If $b=107,$ then we have $1=3a^2-107^2,$ so $a^2=\frac{1+107^2}{3}.$ But $107^2\equiv 2^2\equiv 1(\mathrm{mod}3),$ so the right-hand side is not an integer. Thus, $(a,b)=(6,1)$ is the only possibility. Then the answer is $$a^3-3a{b}^2=6^3-3\cdot 6\cdot 1^2=\boxed{198}.$$