1. Given equation

$\left(P+\frac{a}{V^2}\right)(V-b)=RT$

This is the van der Waals equation form. We need the dimensional formula of $ab^{-1}$.

2. Find dimension of $b$

In the factor $(V-b)$, both terms must have the same dimensions.

So, $[b]=[V]$

Since volume has dimension $[V]=[L^3]$

therefore, $[b]=[L^3]$

3. Find dimension of $a$

In the term $P+\frac{a}{V^2}$

both terms must have the same dimensions. Hence, $\left[\frac{a}{V^2}\right]=[P]$

So, $[a]=[P][V^2]$

Now, $[P]=[ML^{-1}T^{-2}]$ and $[V^2]=[L^3]^2=[L^6]$

Thus, $[a]=[ML^{-1}T^{-2}]\,[L^6]=[ML^5T^{-2}]$

4. Find dimension of $ab^{-1}$

$[ab^{-1}] = \frac{[a]}{[b]}$

Substitute the values: $[ab^{-1}] = \frac{[ML^5T^{-2}]}{[L^3]}=[ML^2T^{-2}]$

5. Match with options

$[ab^{-1}] = [ML^2T^{-2}]$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They match.