16) The answer is (A)

Given $x$ as the known base side and $y$ as the unknown side of the parallelepiped, $h$ as the water level, and $V$ as the volume poured, we have that the unknown base side is $y=\frac{V}{h \cdot x}$, from which $y=\frac{40 \mathrm{~cm}}{5 \mathrm{~cm} \cdot 4 \mathrm{~cm}}=$ $2 \mathrm{~cm}$. We then have $V^{\prime}=x^{\prime} \cdot y \cdot h^{\prime}$, from which $V^{\prime}=6 \mathrm{~cm} \cdot 2 \mathrm{~cm} \cdot 15 \mathrm{~cm}=180 \mathrm{~cm}^{3}$.
16) Pouring $40 \mathrm{~cm}^{3}$ of water into a rectangular parallelepiped container with one side of the base measuring $4 \mathrm{~cm}$, the water level reaches $5 \mathrm{~cm}$. Pouring an unknown amount into another rectangular parallelepiped container that has one side of the base measuring $6 \mathrm{~cm}$ and the other side unchanged, the water reaches a level of $15 \mathrm{~cm}$. How many $\mathrm{cm}^{3}$ of water were poured the second time?
(A) 180
(B) 80
(C) 40
(D) 20
(E) $\frac{80}{9}$.