Ta có:\(\left. {\begin{array}{*{20}{l}}{AC \bot AB}\\{AC \bot SB{\mkern 1mu} {\mkern 1mu} \left( {SB \bot \left( {ABC} \right)} \right)}\end{array}} \right\}\)\( \Rightarrow AC \bot \left( {SAB} \right) \Rightarrow AC \bot SA\)   

\( \Rightarrow \) SA là hình chiếu vuông góc của SC trên (SAB) ⇒\(\widehat {\left( {SC;\left( {SAB} \right)} \right)} = \widehat {\left( {SC;SA} \right)} = \widehat {CSA} = {30^0}\)

\(\left. {\begin{array}{*{20}{l}}{\left( {SAC} \right) \cap \left( {ABC} \right) = AC}\\{\left( {SAC} \right) \supset SA \bot AC}\\{\left( {ABC} \right) \supset AB \bot AC}\end{array}} \right\}\)\( \Rightarrow \widehat {\left( {\left( {SAC} \right);\left( {ABC} \right)} \right)} = \widehat {\left( {SA;AB} \right)} = \widehat {SAB} = {60^0}\)

\(SB \bot \left( {ABC} \right) \Rightarrow SB \bot AB \Rightarrow \Delta SAB\) vuông tại B

\( \Rightarrow AB = SB.cot60 = a.\dfrac{1}{{\sqrt 3 }} = \dfrac{{a\sqrt 3 }}{3}\)

\( \Rightarrow SA = \sqrt {S{B^2} + A{B^2}} {\rm{\; = }}\sqrt {{a^2} + \dfrac{{{a^2}}}{3}} {\rm{\;}} = \dfrac{{2a}}{{\sqrt 3 }}\) 

Xét tam giác vuông SAC ta có: \(AC = SA.\tan 30 = \dfrac{{2a}}{{\sqrt 3 }}.\dfrac{1}{{\sqrt 3 }} = \dfrac{{2a}}{3}\)

⇒ \({S_{ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}\dfrac{{a\sqrt 3 }}{3}.\dfrac{{2a}}{3} = \dfrac{{{a^2}\sqrt 3 }}{9}\)

⇒ \({V_{S.ABC}} = \dfrac{1}{3}SB.{S_{ABC}} = \dfrac{1}{3}.a.\dfrac{{{a^2}\sqrt 3 }}{9} = \dfrac{{{a^3}\sqrt 3 }}{{27}}\)

Chọn A.