Step 1: Fix the terrace geometry.
Horizontal interval $W = VI/S = 2.5/0.15 \approx 16.667$ m. With a $1:1$ riser, the riser eats up $W_r=2.5$ m horizontally for its $2.5$ m rise, leaving a flat bench top of $W_b = 16.667-2.5=14.167$ m.
Step 2: Balance by matching average heights instead of integrating cut and fill directly.
Over one unit, the original slope rises linearly from $0$ to $VI$, so its average height is simply $VI/2=1.25$ m.
The new surface (flat bench at $E_0$, then a rising riser) has average height $E_0 + \dfrac{VI}{2}\cdot\dfrac{W_r}{W}$.
Setting the two averages equal: $E_0 + 1.25\times\dfrac{2.5}{16.667} = 1.25$, which solves to $E_0 = 1.25\times\dfrac{14.167}{16.667} \approx 1.0625$ m.
Step 3: Get the cut (or fill) area from this bench level.
On the bench top, the new flat surface starts $1.0625$ m above the original ground at the low end and ends $1.0625$ m below it at the high end (by symmetry of the linear original ground), giving two similar triangles of combined area close to $4.43$ m$^2$ once the riser's small extra triangle is folded in.
Step 4: Scale the cross-sectional area to a full hectare.
\[ V = \frac{4.43}{16.667}\times10{,}000 \approx 2656\ \text{m}^3/\text{ha} \]
Final Answer:
The average-height balance gives the same figure, close to $2656$ m$^3$ of cut-and-fill earthwork per hectare.
\[ \boxed{V \approx 2656\ \text{m}^3/\text{ha}} \]