Gọi $O=AC\cap BD$

Từ giả thuyết suy ra $A'O\bot \left( ABCD \right)$

${{S}_{ABCD}}=BC.CD.\sin {{120}^{0}}=\frac{{{a}^{2}}\sqrt{3}}{2}$

Vì $\widehat{BCD}={{120}^{0}}$ nên $\widehat{ABC}={{60}^{0}}\Rightarrow \Delta ABC$ đều

$\Rightarrow AC=a\Rightarrow A'O=\sqrt{A'{{A}^{2}}-A{{O}^{2}}}=\sqrt{\frac{49{{a}^{2}}}{4}-\frac{{{a}^{2}}}{4}}=2\sqrt{3}a$

Suy ra ${{V}_{ABCD.A'B'C'D'}}=3{{a}^{3}}$