jmc

Let $\alpha =x+yi$ and $\beta =x-yi.$ Then from $|\alpha -\beta |=2\sqrt{3},$ $2|y|=2\sqrt{3},$ so $|y|=\sqrt{3}.$

Now, $\frac{\alpha }{{\beta }^2}$ is real. Since $\alpha$ and $\beta$ are complex conjugates, ${\alpha }^2{\beta }^2$ is real, so $\frac{\alpha }{{\beta }^2}\cdot {\alpha }^2{\beta }^2={\alpha }^3$ is real. But $${\alpha }^3=(x+yi{)}^3=(x^3-3x{y}^2)+(3x^{2}y-y^3)i,$$so $3x^{2}y-y^3=0,$ or $y(3x^2-y^2)=0.$ Since $|y|=\sqrt{3},$ $y\ne 0,$ so $3x^2=y^2=3,$ which means $x^2=1.$ Therefore, $$|\alpha |=\sqrt{x^2+y^2}=\sqrt{1+3}=\boxed{2}.$$