Lời giải chi tiết:

Trong \(\left( {BCC'B'} \right)\) kẻ \(B'H \bot BC\,\,\left( {H \in BC} \right) \Rightarrow B'H \bot \left( {ABC} \right)\).

Trong \(\left( {ABC} \right)\) kẻ \(HK//AC \Rightarrow HK \bot AB\,\,\left( {K \in AB} \right)\) ta có:

\(\left\{ \begin{array}{l}AB \bot HK\\AB \bot B'H\end{array} \right. \Rightarrow AB \bot \left( {B'HK} \right) \Rightarrow AB \bot B'K\).

\(\begin{array}{l}\left\{ \begin{array}{l}\left( {ABB'A'} \right) \cap \left( {ABC} \right) = AB\\\left( {ABB'A'} \right) \supset B'K \bot AB\\\left( {ABC} \right) \supset HK \bot AB\end{array} \right.\\ \Rightarrow \angle \left( {\left( {ABB'A'} \right);\left( {ABC} \right)} \right) = \angle \left( {B'K;HK} \right) = \angle B'KH = {45^0}\end{array}\)

Đặt \(HK = x \Rightarrow BK = HK.\cot {60^0} = \dfrac{x}{{\sqrt 3 }}\).

Trong tam giác vuông \(B'HK:\,\,HK = B'H = x;\,\,B'K = HK\sqrt 2  = x\sqrt 2 \).

Trong tam giác vuông \(BB'K:\,\,BB{'^2} = B'{K^2} + K{B^2}\)

\( \Leftrightarrow 4{a^2} = 2{x^2} + \dfrac{{{x^2}}}{3} = \dfrac{7}{3}{x^2} \Leftrightarrow x = \dfrac{{2\sqrt {21} }}{7}a = B'H\).

Ta có: \(AB = BC.\cos {60^0} = 2a.\dfrac{1}{2} = a;\,\,AC = BC\sin 60 = 2a.\dfrac{{\sqrt 3 }}{2} = a\sqrt 3 \).

\( \Rightarrow {S_{ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}.a.a\sqrt 3  = \dfrac{{{a^2}\sqrt 3 }}{2}\)

Vậy \({V_{ABC.A'B'C'}} = B'H.{S_{ABC}} = \dfrac{{2\sqrt {21} a}}{7}.\dfrac{{{a^2}\sqrt 3 }}{2} = \dfrac{{3{a^3}}}{{\sqrt 7 }}\).

Chọn B.