Lời giải chi tiết:

* \(\widehat{ABC}={{60}^{0}}\Rightarrow \Delta ABC\) đều

\(\Rightarrow OA=\frac{a}{2};\,\,OB=\frac{a\sqrt{3}}{2}\)

* Vẽ \(OH\bot AB\Rightarrow OH\bot \left( SAB \right)\)

\(\Rightarrow \widehat{\left( SO;\left( SAB \right) \right)}=\widehat{\left( SO;SH \right)}=\widehat{OSH}={{30}^{0}}\)

* Xét tam giác vuông AOB: \(\frac{1}{O{{H}^{2}}}=\frac{4}{{{a}^{2}}}+\frac{4}{3{{a}^{2}}}\Rightarrow OH=\frac{a\sqrt{3}}{4}\)

* Xét tam giác vuông SHO:

\(\sin {{30}^{0}}=\frac{OH}{SO}\Rightarrow SO=\frac{\frac{a\sqrt{3}}{4}}{\frac{1}{2}}=\frac{a\sqrt{3}}{2}\)

Xét tam giác vuông SAO: \(SA=\sqrt{S{{O}^{2}}-A{{O}^{2}}}=\sqrt{\frac{3{{a}^{2}}}{4}-\frac{{{a}^{2}}}{4}}=\frac{a\sqrt{2}}{2}\)   

\({{V}_{S.ABCD}}=\frac{1}{3}SA.2{{S}_{\Delta ABC}}=\frac{1}{3}.\frac{a\sqrt{2}}{2}.2.\frac{1}{2}.a.\frac{a\sqrt{3}}{2}=\frac{{{a}^{3}}\sqrt{6}}{12}\)

Chọn C.