Lời giải chi tiết:

Kẻ \(A'H \bot \left( {ABCD} \right);HM \bot AB;HN \bot AD\)

Ta có: \(\left. \begin{array}{l}A'H \bot AB\\HM \bot AB\end{array} \right\} \Rightarrow AB \bot \left( {A'HM} \right) \Rightarrow AB \bot A'M\)

\(\left. \begin{array}{l}\left( {ABB'A'} \right) \cap \left( {ABCD} \right) = AB\\\left( {ABB'A'} \right) \supset A'M \bot AB\\\left( {ABCD} \right) \supset HM \bot AB\end{array} \right\} \Rightarrow \widehat {\left( {\left( {ABB'A'} \right);\left( {ABCD} \right)} \right)} = \widehat {\left( {A'M;HM} \right)} = \widehat {A'MH} = {45^o}\)

Chứng minh tương tự ta có \(\widehat {A'NH} = {60^0}\)

Đặt \(A'H = x\) khi đó ta có:

\(A'N = \dfrac{x}{{\sin 60}} = \dfrac{{2x}}{{\sqrt 3 }},AN = \sqrt {AA{'^2} - A'{N^2}}  = \sqrt {1 - \dfrac{{4{x^2}}}{3}}  = HM\)

Mà \(HM = x.\cot 45 = x\)

\( \Rightarrow x = \sqrt {1 - \dfrac{{4{x^2}}}{3}}  \Leftrightarrow {x^2} = 1 - \dfrac{{4{x^2}}}{3} \Leftrightarrow \dfrac{{7{x^2}}}{3} = 1 \Rightarrow {x^2} = \dfrac{3}{7} \Rightarrow x = \sqrt {\dfrac{3}{7}} \)

\({S_{ABCD}} = \sqrt 3 .\sqrt 7  = \sqrt {21} \)

Vậy \({V_{ABCD.A'B'C'D'}} = A'H.{S_{ABCD}} = \sqrt {\dfrac{3}{7}} .\sqrt {21}  = 3\)

Chọn A.