\[b_{6} = \frac{1}{27};\ \ \ q = \frac{1}{3}:\]
\[b_{6} = b_{1} \cdot q^{5}\]
\[\frac{1}{3^{3}} = \left( \frac{1}{3} \right)^{5} \cdot b^{1}\]
\[b_{1} = \frac{1}{3^{3}} \cdot 3^{5}\]
\[b_{1} = 3^{2} = 9.\]