Chọn C

Gọi M là trung điểm BC suy ra $AM\perp BC$.

Khi đó $\left\{\begin{array}{l}AM\perp BC\\ AM\perp B{B}^{\text{'}}\end{array}\right.$ nên $AM\perp \left(BC{C}^{\text{'}}{B}^{\text{'}}\right)$ do đó $\left(A{B}^{\text{'}},\left(BC{C}^{\text{'}}{B}^{\text{'}}\right)\right)=\left(A{B}^{\text{'}},M{B}^{\text{'}}\right)=\widehat{A{B}^{\text{'}}M}$.

Theo đề bài, ta có $\widehat{A{B}^{\text{'}}M}=30°, AM=\frac{a\sqrt{3}}{2}$ nên $BM=\frac{AM}{\tan 30°}=\frac{a\sqrt{3}}{2}:\frac{\sqrt{3}}{3}=\frac{3a}{2}$.

Ta có $B{B}^{\text{'}}=\sqrt{B^{\text{'}}{A}^2-B{M}^2}=\sqrt{{\left(\frac{3a}{2}\right)}^2-{\left(\frac{a}{2}\right)}^2}=a\sqrt{2}$.

Thể tích khối lăng trụ ABC.A'B'C' là $V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}=B{B}^{\text{'}}.S_{ABC}=a\sqrt{2}.\frac{a^2\sqrt{3}}{4}=\frac{a^3\sqrt{6}}{4}$.