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

Let us write the unknown complex number in the usual form $$z=x+iy$$, where $$x=\text{Re}(z)$$ and $$y=\text{Im}(z)$$. Its conjugate is therefore $$\overline z=x-iy$$ and, very importantly, $$\text{Re}(z)=\text{Re}(\overline z)=x$$.

Now we list the four given vertices in coordinate form (treating a complex number $$a+ib$$ as the point $$(a,b)$$ in the Argand plane):

$$\begin{aligned} A&:z &&\;\;\;\;=(x,y),\\ B&:\overline z &&\;\;\;\;=(x,-y),\\ C&:\overline z-2\,\text{Re}(\overline z)&&\;\;\;\;=\bigl(x-2x,\,-y\bigr)=(-x,-y),\\ D&:z-2\,\text{Re}(z) &&\;\;\;\;=\bigl(x-2x,\,y\bigr)=(-x,y). \end{aligned}$$

So the four points are $$A(x,y),\;B(x,-y),\;C(-x,-y),\;D(-x,y).$$ One can notice immediately that these points are the four corners of a rectangle centred at the origin; the problem tells us it is actually a square.

We invoke the distance formula between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$:

$$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

Taking consecutive vertices $$A(x,y)$$ and $$B(x,-y)$$, we have

$$AB=\sqrt{(x-x)^2+(-y-y)^2}=\sqrt{0^2+( -2y)^2}=2|y|.$$

Similarly, for the next side $$BC$$ between $$B(x,-y)$$ and $$C(-x,-y)$$ we get

$$BC=\sqrt{(-x-x)^2+(-y+ y)^2}=\sqrt{(-2x)^2+0^2}=2|x|.$$

Because all four sides of a square are equal, we must have

$$2|y|=2|x|\;\;\Longrightarrow\;\;|x|=|y|.$$

The problem states that each side of the square measures 4 units, so

$$2|y|=4\;\;\Longrightarrow\;\;|y|=2.$$

Using $$|x|=|y|$$ immediately gives

$$|x|=2.$$

Finally, the modulus of $$z$$ is obtained from the definition $$|z|=\sqrt{x^2+y^2}:$$

$$|z|=\sqrt{x^2+y^2}=\sqrt{(2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt2.$$

Hence, the correct answer is Option 3.