Answer: 198

Small Hint:

Expand $(a+bi)^3$ and set its imaginary part equal to $107$

Big Hint:

The resulting equation shows that the positive integer $b$ divides $107$

Solution:

Expanding and using that $c$ is real gives $$b(3a^2-b^2)=107.$$ Since $107$ is prime, $b=1$ or $107.$ The latter would require $3a^2=11450,$ which is impossible. Thus $b=1,$ and $3a^2-1=107,$ so $a=6.$ The real part is $$c=a^3-3ab^2=216-18=198.$$