\[\boxed{\mathbf{885\ (885).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[b_{1} = 2\sqrt{3};\ \ q = \sqrt{3}\]
\[b_{1}^{2} + b_{2}^{2} + b_{3}^{2} + b_{4}^{2} + b_{5}^{2} + b_{6}^{2} =\]
\[= b_{1}^{2} + b_{1}^{2}q^{2} + b_{1}^{2q^{4}} + b_{1}^{2q^{6}} +\]
\[+ b_{1}^{2} \cdot q^{8} + b_{1}^{2} \cdot q^{10}\ \]
\[b_{1}^{2} = \left( 2\sqrt{3} \right)^{2} = 4 \cdot 3 = 12,\]
\[\ \ q = \frac{b_{1}^{2}q^{2}}{b_{1}²} = q² = \left( \sqrt{3} \right)^{2} = 3\]
\[S_{6} = \frac{12 \cdot \left( 3^{6} - 1 \right)}{3 - 1} =\]
\[= 6 \cdot (729 - 1) = 6 \cdot 728 =\]
\[= 4368.\]
\[Ответ:4368.\]