The solution is AM=6 (as shown in the figure), and the slope of EF after reinforcement for the same reason can be marked as tan ∠ f = en: fn =1:1.4, and then FN=7 can be calculated.
Because MN=ED= 1 and AN=AM-MN=6- 1=5 after double height, then AF=FN-AN=7-5=2.
How much earthwork is needed is to find the volume of the reinforced part, and the cross-sectional area of the reinforced part is the area of trapezoidal EDAF S=(ED+AF)*DM÷2, so S=7.5, then V=Sh=7.5×4000=30000 (square).