\[b_{5} = 27;\ \ q = \sqrt{3}:\]
\[b_{5} = b_{1} \cdot q^{4}\]
\[b_{1} = \frac{b_{5}}{q^{4}} = \frac{27}{\left( \sqrt{3} \right)^{4}} = \frac{27}{9} = 3.\]
\[Ответ:3.\]