根据《钢标》8.3.2，$K_b=\frac{\sum(I_{bi}/l_i)}{I_c/H}=\frac{4.56\times10^9/206\times10^3}{9.41\times10^8/13000}=0.306$，$\alpha_K=1$，$\gamma=N_1/N_2=108/300=0.36$，$\alpha_N=1+\frac{H_1}{H_2}\frac{(\gamma-0.2)}{1.2}=1+\frac{4400}{8600}\frac{(0.36-0.2)}{1.2}=1.068$，$H_0=\alpha_N\left[\sqrt{\frac{4+7.5K_b}{1+7.5K_b}}-\alpha_K\left(\frac{H_1}{H}\right)^{1+0.8K_b}\right]H=1.068\times\left[\sqrt{\frac{4+7.5\times0.306}{1+7.5\times0.306}}-1\times\left(\frac{4400}{13000}\right)^{1+0.8\times0.306}\right]\times13000=15630\mathrm{mm}$，选B。