Given, the roots of ${\left(z-4\right)}^3=8i$ are $a-2i, b+i$ and $c+i.$

Let $z=x+iy,$ then ${\left(x+iy-4\right)}^3=8i$

$\Rightarrow {\left(x-4+iy\right)}^3=8i$

$\Rightarrow {\left(x-4\right)}^3+{\left(iy\right)}^3+3{\left(x-4\right)}^2\left(iy\right)+3\left(x-4\right){\left(iy\right)}^2=8i$

Using $i^2=-1, i^3=-i,$

$\Rightarrow {\left(x-4\right)}^3-i{y}^3+3{\left(x-4\right)}^2\left(iy\right)-3\left(x-4\right)y^2=8i$

On comparing the real and imaginary parts on both sides of the equation, we get

${\left(x-4\right)}^3-3\left(x-4\right)y^2=0 ...\left(i\right)$ and

 $-y^3+3{\left(x-4\right)}^{2}y=8$

$y\left[-y^2+3{\left(x-4\right)}^2\right]=8 ...\left(ii\right)$

From the equation $\left(i\right),$

${\left(x-4\right)}^3-3\left(x-4\right)y^2=0$

$\Rightarrow \left(x-4\right)\left[{\left(x-4\right)}^2-3y^2\right]=0$

$\Rightarrow \left(x-4\right)=0$ or ${\left(x-4\right)}^2-3y^2=0$

$\Rightarrow x=4$ or ${\left(x-4\right)}^2=3y^2 ...\left(iii\right)$

On putting, these values in the equation $\left(ii\right),$ we get

$y\left[-y^2+3{\left(4-4\right)}^2\right]=8$ or $y\left[-y^2+3\left(3y^2\right)\right]=8$

$\Rightarrow -y^3=8$ or $y\left[-y^2+9y^2\right]=8$

$\Rightarrow y=-2$ or $8y^3=8, \Rightarrow y=1.$

Thus, for $x=4, y=-2.$

Now, for $y=1,$ from equation $\left(iii\right),$

${\left(x-4\right)}^2=3$

$\Rightarrow \left(x-4\right)=\pm \sqrt{3}$

$\Rightarrow x=4\pm \sqrt{3}.$

 Hence, the roots of the given equation are $4-2i, 4\pm \sqrt{3}+i.$

And, the given roots are $a-2i, b+i$ and $c+i.$

On comparing, we get $\sqrt{abc}=\sqrt{4\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)}$

$\Rightarrow \sqrt{abc}=\sqrt{4\left(16-3\right)}$

$\Rightarrow \sqrt{abc}=2\sqrt{13}.$