The dimensions $[MLT^{-2}A^{-2}]$ belong to the:

Let us determine the dimensional formulae for the given physical quantities:

1. Magnetic permeability ($\mu_0$): The force per unit length between two parallel current-carrying conductors is given by $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$. Rearranging for $\mu_0$, we get $\mu_0 = \frac{F \cdot 2\pi d}{I_1 I_2 l}$. Substituting the dimensions: Force $[F] = [MLT^{-2}]$, distance $[d] = [L]$, length $[l] = [L]$, and current $[I] = [A]$. Dimensions of $\mu_0 = \frac{[MLT^{-2}][L]}{[A]^2 [L]} = [MLT^{-2}A^{-2}]$.

2. Electric permittivity ($\varepsilon_0$): From Coulomb's law, $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \implies \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. Its dimensions are $\frac{[AT]^2}{[MLT^{-2}][L]^2} = [M^{-1}L^{-3}T^4A^2]$.

3. Magnetic flux ($\Phi_B$): $\Phi_B = B \cdot A$. The dimensions of magnetic field $[B] = [MT^{-2}A^{-1}]$ and area $[A] = [L^2]$. So, dimensions of $\Phi_B = [ML^2T^{-2}A^{-1}]$.

4. Self-inductance ($L$): Energy stored in an inductor is $U = \frac{1}{2}LI^2 \implies L = \frac{2U}{I^2}$. Its dimensions are $\frac{[ML^2T^{-2}]}{[A]^2} = [ML^2T^{-2}A^{-2}]$.

Therefore, the dimensions $[MLT^{-2}A^{-2}]$ correspond to magnetic permeability .