The value of $b$ occurs when the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ degenerates to an isosceles triangle. This occurs when the altitude from $A$ to $B C D$ is parallel to the plane. Let $s=A B$. Then the altitude from $A$ intersects the center $E$ of face $B C D$. Since $E B=\frac{s}{\sqrt{3}}$, it follows that $A^{\prime} C^{\prime}=A E=\sqrt{s^{2}-\frac{s^{2}}{3}}=\frac{s \sqrt{6}}{3}$. Then since $B D$ is parallel to the plane, $B^{\prime} D^{\prime}=s$. Then the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $4=\frac{1}{2} \cdot \frac{s^{2} \sqrt{6}}{3}$, implying $s^{2}=4 \sqrt{6}$, or $s=2 \sqrt[4]{6}$.