If a and b are positive integer such that $N={(a+ib)}^3-107i$ is a positive integer. Then the value of N is

$N=(a+ib{)}^3-107i$

$=\left(a^3-3a{b}^2\right)+i\left[3a^{2}b-b^3\right]-107i$

Since N is a positive integer.

$3a^{2}b-b^3-107=0$

$\text{ Or }b\left(3a^2-b^2\right)=107$

$\Rightarrow b=1\text{ and }3a^2-b^2=107\text{ (as }107\text{ is prime) }$

$\therefore a=6$

$\text{ Or }b=107\text{ and }3a^2-(107{)}^2=1$

So, a is not integer which is not possible

$\therefore a=6\text{ and }b=1$

$\therefore N=216-3\times 6=216-18=198$