Inspiring

From the expression, $|z|^3$ is real, $2|z|$ is real and $z$ is complex. As the complex part of the RHS = complex part of the LHS, $img(|z|^3 + 2|z| + z) = img(z) = img(68994 + 40i) = 40 \\ img(z) = 40$

Now to find the real part, let $z = x + 40i$ such that $|z| = \sqrt{x^2 + 40^2} = \sqrt{x^2 + 1600}$ . Using $|z|^3 = |z||z|^2$ , The given equation now becomes $\sqrt{x^2 + 1600}(x^2 + 1600) + 2\sqrt{x^2 + 1600} + x = 68994 \\ \sqrt{x^2 + 1600}(x^2 + 1602) + x = 68994$

By trying out numbers (which was influenced by trying to make the surds into integers) I noticed by letting $x = \pm 9$ , the surd terms became $\sqrt{(\pm9)^2 + 1600} = \sqrt{1681} = 41$ so by substituting $x = \pm9$ into the equation, the the result was $41 * 1683 \pm 9 = 68994$ which is true for the negative value of $\pm9$ so $x = -9$ is a solution and by extension, $z = -9 + 40i$

The final solution is now $3|\mathord{-} 9| + 4 = \boxed{31}$

let z=a+ib and |z|=(a^2+b^2)^(1/2) so comparing real and img. part we have b=40 and thus (a^2+1600)^(3/2)+2((a^2+1600)^(1/2))+a=68994 on solving this we have a=-9 so answer=3x|-9|+4=31