Chọn A

Gọi M, N lần lượt là trung điểm của B'C', BC.

$\widehat{A\text{'}BA}=\widehat{C\text{'}A\text{'}C}={90}^{\circ }\Rightarrow A\text{'}B=A\text{'}C\Rightarrow A\text{'}N\perp BC\left(1\right)$.

Theo bài ra $\widehat{BAD}={120}^{°}\Rightarrow \Delta A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}},\Delta ABC$ đều $\Rightarrow BC\perp AN\left(2\right)$.

Từ $\left(1\right),\left(2\right)\Rightarrow BC\perp \left(AA\text{'}MN\right)\Rightarrow \left(AA\text{'}MN\right)\perp \left(BCC\text{'}B\text{'}\right)$

Kẻ $A^{\text{'}}P\perp \left(B{B}^{\text{'}}{C}^{\text{'}}C\right)\Rightarrow P\in MN$. Gọi Q là hình chiếu vuông góc của A' lên BB'.

$$\begin{gathered} \Rightarrow \left(\left(A^{\text{'}}AD\right),\left(A{A}^{\text{'}}{B}^{\text{'}}B\right)\right)=\left(\left(B{B}^{\text{'}}{C}^{\text{'}}C\right),\left(A{A}^{\text{'}}{B}^{\text{'}}B\right)\right)=\widehat{A^{\text{'}}QP}=\alpha . \\ \tan \alpha =\frac{A^{\text{'}}P}{QP}=\sqrt{2}\Rightarrow A^{\text{'}}P=\frac{a\sqrt{2}}{2}.\Rightarrow A^{\text{'}}Q=\frac{a\sqrt{3}}{2}\Rightarrow A^{\text{'}}B=a\sqrt{3} \\ \Rightarrow A^{\text{'}}B=a\sqrt{3}\Rightarrow B{B}^{\text{'}}=2a \end{gathered}$$