Gọi N là trung điểm của A′B′, khi đó: \(MN//A'B\).

Suy ra: \(\left\{ \begin{array}{l}MN = \frac{{A'B}}{2} = 2a\\(A\prime B,CM) = (NM,CM) = \widehat {CMN} = 30^\circ \end{array} \right.\) .

Ta có \({S_{CMN}} = \frac{1}{2}.CM.MN.\sin \widehat {CMN} = \frac{1}{2}.a\sqrt 2 .2a.\sin 30^\circ = \frac{{\sqrt 2 }}{2}{a^2}\).

Vì \(A'B//\left( {CMN} \right)\)nên \(d\left( {A'B,CM} \right) = d\left( {A'B,\left( {CMN} \right)} \right) = d\left( {B,\left( {CMN} \right)} \right) = a\).

Khi đó: \[{V_{B.CMN}} = \frac{1}{3}d\left( {B,\left( {CMN} \right)} \right).{S_{CMN}} = \frac{1}{3}a.\frac{{{a^2}\sqrt 2 }}{2} = \frac{{{a^3}\sqrt 2 }}{6}\] (∗)

Ta có: \[{S_{BMN}} = \frac{1}{2}{S_{BNB\prime }} = \frac{1}{4}{S_{A\prime BB\prime }} \Rightarrow {V_{B.CMN}} = {V_{C.BMN}} = \frac{1}{4}{S_{C.A\prime BB\prime }} = \frac{1}{4}.\frac{1}{3}.{V_{ABC.A\prime B\prime C\prime }}\].

\[ \Rightarrow {V_{ABC.A\prime B\prime C\prime }} = 12{V_{B.CMN}} = 12.\frac{{{a^3}\sqrt 2 }}{6} = 2\sqrt 2 {a^3}\] . Chọn A.