根据《地处规》7.1.6，$f_{cu}\ge4\frac{\lambda R_a}{A_p}[1+\frac{\gamma_m(d-0.5)}{f_{spa}}]$。首先计算$\gamma_m$：$\gamma_m=\frac{18.6\times1+18.9\times0.8}{1.8}=18.73\mathrm{kN/m^3}$。$f_{spa}$同题12为250kPa。$A_p=0.1256\mathrm{m^2}$，$R_a=680\mathrm{kN}$，$\lambda=0.9$，代入得$f_{cu}\ge4\times\frac{0.9\times680}{0.1256}\times[1+\frac{18.73\times(1.8-0.5)}{250}]=4\times4874.5\times[1+\frac{24.35}{250}]=19498\times(1+0.0974)=19498\times1.0974=21400\mathrm{kPa}=21.4\mathrm{MPa}$，选D。