根据《钢标》8.2.1，$\lambda_y=4500/73.5=61.2$，$\beta_{tx}=1.0$，$b/h=300/390=0.77<0.8$，为b类，查表D.0.2得$\varphi_y=0.8$，$\varphi_b=1.07-61.2^2/44000=0.985$，$\frac{N}{\varphi_y A f}+\eta\frac{\beta_{tx} M_x}{\varphi_b W_{1x} f}=\frac{896×10^3}{0.8×133.3×10^2×215}+1×\frac{1×224×10^6}{0.985×1940×10^3×215}=0.391+0.545=0.936$，选A。