The five unit squares have total area $5$, so each piece must have area $\frac52$. Treating the cut point as $(c,0)$, the shaded region (bounded by the cut, then the pentomino's own boundary through $(3,3),(3,1),(2,1),(2,0)$, then back along the bottom edge to $(c,0)$) is the polygon with vertices $(c,0),(3,3),(3,1),(2,1),(2,0)$.
- By the shoelace theorem, its area is $$\begin{aligned} &\frac12\big|c(3)-3(0)+3(1)-3(3)+3(1)-2(1)+2(0)-2(1)+2(0)-c(0)\big|\\ &=\frac12|3c-7|=\frac{7-3c}2 \end{aligned}$$ since $3c<7$.
- Setting $\dfrac{7-3c}2=\dfrac52$ gives $7-3c=5$, so $c=\boxed{\dfrac23}$.
- Answer is (C).
