Chọn C

Ta có

$\left.\begin{array}{l}\left(SAB\right)\perp \left(ABC\right)\\ \left(SAC\right)\perp \left(ABC\right)\\ \left(SAB\right)\cap \left(SAC\right)=SA\end{array}\right\}\Rightarrow SA\perp \left(ABC\right)\Rightarrow V=\frac{1}{3}.SA.S_{\Delta ABC}$

Trong mặt phẳng $\left(ABC\right)$ kẻ $AI\perp BC\Rightarrow SI\perp BC$

Suy ra góc giữa 2 mặt phẳng $\widehat{\left(\left(SBC\right),\left(ABC\right)\right)}=\left(AI,SI\right)=\widehat{SIA}={60}^0$

Trong tam giác ABC ta có

$B{C}^2=A{B}^2+A{C}^2-2.AB.AC.\cos A=a^2+4a^2-2.a.2a.\left(-\frac{1}{2}\right)=7a^2\Rightarrow BC=a\sqrt{7}$

Ta có $S_{\Delta ABC}=\frac{1}{2}AB.AC.\sin A=\frac{1}{2}.a.2a.\sin {120}^0=\frac{a^2\sqrt{3}}{2}$

Mà $S_{\Delta ABC}=\frac{1}{2}AI.BC\Rightarrow AI=\frac{2S_{\Delta ABC}}{BC}=\frac{2.\frac{a^2\sqrt{3}}{2}}{a\sqrt{7}}=\frac{a\sqrt{21}}{7}$

Trong tam giác $\left(SAI\right)$ vuông tại A ta có $\tan \widehat{SIA}=\frac{SA}{AI}\Rightarrow SA=AI.\tan {60}^0=\frac{a\sqrt{21}}{7}.\sqrt{3}=\frac{3a\sqrt{7}}{7}$

Vậy thể tích của khối chóp là $V=\frac{1}{3}.SA.S_{\Delta ABC}=\frac{1}{3}.\frac{3a\sqrt{7}}{7}.\frac{a^2\sqrt{3}}{2}=\frac{a^3\sqrt{21}}{14}.$