Cho hình chóp S.ABCD có đáy ABCD là hình chữ nhật, SA vuông góc (ABCD)

Gọi M là trung điểm của SD ta có $AG\cap \left(SBD\right)=\left\{M\right\}$ nên $\frac{d\left(G;\left(SBD\right)\right)}{d\left(A;\left(SBD\right)\right)}=\frac{GM}{AM}=\frac{1}{3}.$

$\Rightarrow d\left(G;\left(SBD\right)\right)=\frac{1}{3}d\left(A;\left(SBD\right)\right).$

Trong (ABCD) kẻ $AH\perp BD,$ trong (SAH) kẻ $AK\perp SH$

Ta có

$\left\{\begin{array}{l}BD\perp AH\\ BD\perp SA\end{array}\right.\Rightarrow BD\perp \left(SAH\right)\Rightarrow BD\perp AK$

$\left\{\begin{array}{l}AK\perp BD\\ AK\perp SH\end{array}\right.\Rightarrow AK\perp \left(SBD\right)$

$\Rightarrow d\left(A;\left(SBD\right)\right)=AK.$

Ta có: $AH=\frac{AB.AD}{\sqrt{A{B}^2+A{D}^2}}=\frac{a.2a}{\sqrt{a^2+4a^2}}=\frac{2a}{\sqrt{5}}.$

$\Rightarrow AK=\frac{SA.AH}{\sqrt{S{A}^2+A{H}^2}}=\frac{a.\frac{2a}{\sqrt{5}}}{\sqrt{a^2+\frac{4a^2}{5}}}=\frac{2a}{3}.$

Vậy $d\left(G;\left(SBD\right)\right)=\frac{1}{3}d\left(A;\left(SBD\right)\right)=\frac{1}{3}.\frac{2a}{3}=\frac{2a}{9}.$

Chọn B.