根据《混规》第7.1.2条

$$ \rho_ {\mathrm{te}} = \frac {A _ {\mathrm{s}}}{A _ {\mathrm{te}}} = \frac {8 \times 2 0 1 . 1}{3 0 0 \times 3 0 0} = 0. 0 1 7 9 > 0. 0 1 $$

$$ \psi = 1. 1 - 0. 6 5 \frac {f _ {\mathrm{tk}}}{\rho_ {\mathrm{te}} \sigma_ {\mathrm{s}}} = 1. 1 - 0. 6 5 \times \frac {2 . 0 1}{0 . 0 1 7 9 \times 1 7 0} = 0. 6 7 \begin{array}{l} < 1. 0 \\ > 0. 2 \end{array} $$

根据《混规》表7.1.2-1，轴心受拉， $\alpha_{\mathrm{cr}} = 2.7$

$$ \omega_ {\max} = \alpha_ {\mathrm{cr}} \psi \frac {\sigma_ {\mathrm{s}}}{E _ {\mathrm{s}}} \left(1. 9 c _ {\mathrm{s}} + 0. 0 8 \frac {d _ {\mathrm{eq}}}{\rho_ {\mathrm{te}}}\right) = 2. 7 \times 0. 6 7 \times \frac {1 7 0}{2 \times 1 0 ^ {5}} \times \left(1. 9 \times 3 0 + 0. 0 8 \times \frac {1 6}{0 . 0 1 7 9}\right) = 0. 1 9 8 \mathrm{mm} $$

因此选（B）。