If the four complex numbers $z,\overline{z},\overline{z}-2\mathrm{Re}\left(\overline{z}\right)$ and  $z-2\mathrm{Re}\left(z\right)$ represent the vertices of a square of side 4 units in the Argand plane, then $\left|z\right|$ is equal to

Let  $z=z+iy$ 

So, vertices of square are $(x,y),(x,-y),(-x,-y)\text{ and }(-x,y)$ Clearly, origin is the point of intersection of diagonals of the square.

Length of sided of square is 4 units.

So, length of diagonal is $4\sqrt{2}\text{ units. }$

Therefore, distance of each vertex from origin is $2\sqrt{2}.$

$\therefore \left|z\right|=2\sqrt{2}$