1. Given dimension

We need the unit corresponding to $\frac{M L^2}{Q^2}$ where $Q$ is electric charge.

Since electric charge has dimension $[Q] = [IT],$ we get $\frac{M L^2}{Q^2} = \frac{M L^2}{I^2 T^2}.$

2. Now find the dimensions of each option

Option A: $\dfrac{Wb}{m^2}$

We know $1\,Wb = 1\,T\cdot m^2$ and tesla has dimension $[T] = \frac{N}{A\,m} = \frac{M L T^{-2}}{I L} = M T^{-2} I^{-1}.$ So, $[Wb] = [T][m^2] = M L^2 T^{-2} I^{-1}.$ Therefore, $\left[\frac{Wb}{m^2}\right] = M T^{-2} I^{-1}.$ This is not equal to $\dfrac{M L^2}{I^2 T^2}$.

So, A is incorrect.

Option B: Henry $(H)$

Inductance is defined by $V = L\frac{dI}{dt}$ so $[L] = \frac{[V][t]}{[I]}.$ Now, $[V] = \frac{[\text{energy}]}{[\text{charge}]} = \frac{M L^2 T^{-2}}{I T} = M L^2 T^{-3} I^{-1}.$ Thus, $[H] = \frac{(M L^2 T^{-3} I^{-1})(T)}{I} = M L^2 T^{-2} I^{-2}.$ Since $Q = IT$, $M L^2 T^{-2} I^{-2} = \frac{M L^2}{Q^2}.$ So, B matches exactly.

Thus, B is correct.

Option C: $\dfrac{H}{m^2}$

Using $[H] = M L^2 T^{-2} I^{-2},$ we get $\left[\frac{H}{m^2}\right] = M T^{-2} I^{-2}.$ This is not equal to $\dfrac{M L^2}{I^2 T^2}$.

So, C is incorrect.

Option D: Weber $(Wb)$

From above, $[Wb] = M L^2 T^{-2} I^{-1}.$ This is not equal to $\dfrac{M L^2}{I^2 T^2}$.

So, D is incorrect.

3. Final answer

The required unit is $\boxed{\text{Henry }(H)}$

4. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So, the answer agrees with the stored correct answer.