4. $\frac{13}{12} \sqrt{2} a^{3}$.
(Method 1) As shown in Figure 2, in the rectangular prism $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, $A B=B C=2 a$, and the height is $x$. Then, $x=\sqrt{a^{2}-\left(\sqrt{2} \cdot \frac{a}{2}\right)^{2}}=\frac{\sqrt{2}}{2} a$. The required volume is
$$
(2 a)^{2} x-4 \times\left(\frac{1}{2} \cdot \frac{a}{2} \cdot x \cdot 2 a\right)+4 \times\left(\frac{1}{6} \cdot \frac{a}{2} \cdot x \cdot \frac{a}{2}\right)=x \cdot \frac{13}{6} a^{2}=\frac{13}{12} \sqrt{2} a^{3} .
$$
(Method 2) The tetrahedron can be divided into the following 3 parts:
(1) 1 square prism with a base side length of $a$ and a height of $\frac{\sqrt{2}}{2} a$, its volume is $a \times a \times \frac{\sqrt{2}}{2} a=\frac{\sqrt{2}}{2} a^{3}$;
(2) 4 right triangular prisms with a base of right angles with legs of $\frac{1}{2} a$ and $\frac{\sqrt{2}}{2} a$, and a height of $a$, the volume of each right triangular prism is $\frac{1}{2} \times \frac{1}{2} a \times \frac{\sqrt{2}}{2} a \times a=\frac{\sqrt{2}}{8} a^{3}$;
(3) 4 triangular pyramids with a base of isosceles right triangles with legs of $\frac{1}{2} a$ and a height of $\frac{\sqrt{2}}{2} a$, the volume of each triangular pyramid is $\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} a \times \frac{1}{2} a \times \frac{\sqrt{2}}{2} a=\frac{\sqrt{2}}{48} a^{3}$.
Therefore, the volume of the hexahedron is $\frac{\sqrt{2}}{2} a^{3}+4 \times \frac{\sqrt{2}}{8} a^{3}+4 \times \frac{\sqrt{2}}{48} a^{3}=\frac{13}{12} \sqrt{2} a^{3}$.
4. Using 4 isosceles trapezoid cardboard pieces with leg lengths of $a$, and top and bottom base lengths of $a$ and $2a$ respectively, and two rectangular cardboard pieces with lengths and widths of $2a$ and $a$ respectively, a hexahedron can be formed. The volume of this hexahedron is $\qquad$