根据《混规》9.7.2条，$N=F\cos45^\circ=\frac{\sqrt{2}}{2}F$，$V=F\sin45^\circ=\frac{\sqrt{2}}{2}F$，$M=F\sin45^\circ\times0.08=0.04\sqrt{2}F$。$M=0.04\sqrt{2}F<0.4Nz=0.4\times\frac{\sqrt{2}}{2}F\times0.25=0.05\sqrt{2}F$，取$M=0.4Nz$。$\alpha_v=(4.0-0.08d)\sqrt{\frac{f_c}{f_y}}=(4.0-0.08\times16)\times\sqrt{\frac{14.3}{300}}=0.594<0.7$。$A_s\ge\frac{V-0.3N}{\alpha_r\alpha_v f_y}+\frac{M-0.4Nz}{1.3\alpha_r\alpha_b f_y z}$，$6\times201.1\ge\frac{\frac{\sqrt{2}}{2}F-0.3\times\frac{\sqrt{2}}{2}F}{0.9\times0.594\times300}+0$，得$F\le390.9\mathrm{kN}$。选C。