Lời giải chi tiết:

+) Vẽ \(SI\bot \left( ABC \right)\Rightarrow \) I là tâm đường tròn nội tiếp D ABC.

+) Vẽ \(IM\bot AC\Rightarrow \widehat{\left( \left( SAC \right);\left( ABC \right) \right)}=\widehat{M}={{60}^{o}}\)

+) Ta có: \(B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB.AC.cos{{120}^{o}}\Rightarrow B{{C}^{2}}=3{{a}^{2}}\Rightarrow BC=a\sqrt{3}\)

+) Nửa chu vi tam giác ABC: \(p=\frac{a+a+a\sqrt{3}}{2}=\frac{a\left( \sqrt{3}+2 \right)}{2}\)

\(\begin{align} & +)\,\,{{S}_{\Delta ABC}}=\frac{1}{2}.a.a.\sin {{120}^{o}}=\frac{1}{2}{{a}^{2}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{3}{{a}^{2}}}{4} \\ & +)\,\,r=IM=\frac{{{S}_{\Delta ABC}}}{p}=\frac{\frac{\sqrt{3}{{a}^{2}}}{4}}{\frac{a\left( \sqrt{3}+2 \right)}{2}}=\frac{\sqrt{3}a}{2\left( \sqrt{3}+2 \right)} \\ & +)\,\,h=SI=IM.\tan {{60}^{o}}=r.\tan {{60}^{o}}=\frac{\sqrt{3}a}{2\left( \sqrt{3}+2 \right)}.\sqrt{3}=\frac{3a}{2\left( \sqrt{3}+2 \right)} \\ & +)\,{{V}_{SABC}}=\frac{1}{3}.h.{{S}_{\Delta ABC}}=\frac{1}{3}.\frac{3a}{2\left( \sqrt{3}+2 \right)}.\frac{\sqrt{3}{{a}^{2}}}{4}=\frac{\sqrt{3}{{a}^{3}}}{8\left( \sqrt{3}+2 \right)} \\\end{align}\)

Chọn A