Why is the Van der Waals constant of water lesser than benzene?
The parameter $a$ from the van der Waals equation

$$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$

is not to be taken only as the measure of strength of intermolecular bonding.

Rather, it is overall measure of significance and frequency of intermolecular interaction.

It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved.

The probability 2 benzene molecules interact is bigger than for 2 water molecules.

Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$. It has bigger value than benzene, in spite of the weaker interaction due the lack of $\pi$ bonding. But its linear molecule is stratched with all the hydrogen atoms in many directions, so collisions are more probable.