根据《地处规》7.1.5及7.1.6，$R_a=u_p\sum q_{si}l_{pi}+\alpha_p q_p A_p=0.4\pi\times(20\times6+10\times3+30\times6+40\times1)+1.0\times450\times0.2^2\pi=521.24\mathrm{kN}$，$R_a=\frac{A_p f_{cu}}{4\lambda}=\frac{15000\times0.2^2\pi}{4\times0.8}=588.75\mathrm{kN}$，取$R_a=521.24\mathrm{kN}$。$f_{spk}=\lambda m\frac{R_a}{A_p}+\beta(1-m)f_{sk}=0.8\times m\times\frac{521.24}{0.2^2\pi}+0.9\times(1-m)\times100=250$，解得$m=0.05$，$m=\frac{d^2}{d_e^2}=\frac{0.4^2}{(1.05s)^2}=0.05$，$s=1.7\mathrm{m}$，选B。