In the van der Waals equation, the term $\dfrac{an^2}{V^2}$ must have units of pressure.

So: $a = \dfrac{P \cdot V^2}{n^2}$

Units of $a = \dfrac{\text{atm} \cdot (\text{dm}^3)^2}{(\text{mol})^2} = \mathbf{\text{atm dm}^6\ \text{mol}^{-2}}$

The constant $b$ has units of dm$^3$ mol$^{-1}$ (it represents excluded molar volume). 'a' corrects for intermolecular attractions.