Lời giải chi tiết:

* Vẽ \(SH\bot BC\Rightarrow SH\bot \left( ABC \right)\)

* Tam giác vuông SHB : \(\sin {{30}^{0}}=\frac{SH}{2a\sqrt{3}}\Rightarrow SH=a\sqrt{3}\Rightarrow BH=\sqrt{S{{B}^{2}}-S{{H}^{2}}}.SB=3a\Rightarrow HC=4a-3a=a\)

* Ta có \(\frac{CB}{CH}=\frac{d\left( B;\left( SAC \right) \right)}{d\left( H;\left( SAC \right) \right)}=4\Rightarrow d\left( B;\left( SAC \right) \right)=4d\left( H;\left( SAC \right) \right)=4HK\)

* Tính HK : \({{\Delta }_{v}}HEC\backsim {{\Delta }_{v}}ABC\Rightarrow \frac{HE}{AB}=\frac{CH}{AC}\Rightarrow HE=\frac{CH.AB}{\sqrt{A{{B}^{2}}+B{{C}^{2}}}}=\frac{a.3a}{5a}=\frac{3a}{5}\)

* Tam giác vuông SHE : \(\frac{1}{H{{K}^{2}}}=\frac{1}{S{{H}^{2}}}+\frac{1}{H{{E}^{2}}}=\frac{1}{3{{a}^{2}}}+\frac{25}{9{{a}^{2}}}=\frac{28}{9{{a}^{2}}}\Rightarrow HK=\frac{3a}{2\sqrt{7}}\)

Đáp số \(\frac{6a}{\sqrt{7}}\)

Chọn đáp án A.