Given: $z^2+i\overline{z}=0$

$\Rightarrow z^2=-i\overline{z}$

Now taking $z=x+iy$ we get,

$\Rightarrow x^2-y^2+2ixy=-ix-y$

Now on comparing both side we get,

$\Rightarrow x^2-y^2=-y \& 2xy=-x$

$\Rightarrow x^2-y^2=-y \& x\left(2y+1\right)=0$

$\Rightarrow x^2-{\left(\frac{-1}{2}\right)}^2=-\left(\frac{-1}{2}\right) \& y=\frac{-1}{2}$

$\Rightarrow x^2=\left(\frac{3}{4}\right) \& y=\frac{-1}{2}$

$\Rightarrow x=\pm \frac{\sqrt{3}}{2} \& y=\frac{-1}{2}$

Hence, $z=\pm \frac{\sqrt{3}}{2} +i\left(\frac{-1}{2}\right)$

$\Rightarrow z^2=\frac{3}{4} -\frac{1}{4}\pm 2i\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$

$\Rightarrow z^2=\frac{1}{2} \pm i\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow \left|z^2\right|={\left(\sqrt{\frac{1}{4}+\frac{3}{4}}\right)}^2=1$