Three vessels, as shown in the figure, possess identical base...

Let the identical base area of the three vessels be $A$ and the equal volume of liquid poured in each be $V$.

The height of the liquid in each vessel depends on its shape:
For vessel A (wider at the top), the cross-sectional area increases with height. Thus, the liquid height $h_A < \dfrac{V}{A}$.
For vessel B (straight walls), the cross-sectional area is constant. Thus, the liquid height $h_B = \dfrac{V}{A}$.
For vessel C (narrower at the top), the cross-sectional area decreases with height. Thus, the liquid height $h_C > \dfrac{V}{A}$.

Therefore, the heights of the liquid columns are in the order: $h_A < h_B < h_C$.

The pressure at the base of a vessel is given by $P = \rho g h$, where $\rho$ is the density of the liquid and $h$ is the height of the liquid column.

The force acting on the base is $F = P \times A = \rho g h A$.

Since the density $\rho$, acceleration due to gravity $g$, and base area $A$ are constant for all three vessels, the force on the base is directly proportional to the height of the liquid ($F \propto h$).

As $h_C$ is the maximum, the force acting on the base will be maximum in vessel C.