Ta có

$\begin{align} & BD=\sqrt{A{{B}^{2}}+A{{D}^{2}}-2AB.AD\cos A}=a\sqrt{3} \\ & AO=\sqrt{\frac{A{{B}^{2}}+A{{D}^{2}}}{2}-\frac{B{{D}^{2}}}{4}}=a\frac{\sqrt{7}}{2}\to AC=a\sqrt{7} \\ \end{align}$

$\to SA=a\sqrt{21}$

Mà ${{S}_{ABC}}=\frac{1}{2}AB.AD\sin A=\frac{{{a}^{2}}\sqrt{3}}{2}$ do đó ${{S}_{ABCD}}={{a}^{2}}\sqrt{3}$.

Vậy $\frac{V}{{{a}^{3}}}=\frac{1}{3}SA.{{S}_{ABC}}=\sqrt{7}$