To solve this problem, we need to determine which physical quantity has the dimensions $[MLT^{-2}A^{-2}]. \\$Let's analyze the dimensions of each option:$\\$1. Magnetic flux $(\Phi): \\$The formula for magnetic flux is $\Phi = B \cdot A,$ where $B$ is magnetic field and $A$ is area.$\\$The dimensions of magnetic field $(B)$ are $[MT^{-2}A^{-1}],$ and area $(A)$ has dimensions $[L^2]. \\$Thus, the dimensions of magnetic flux are $[MT^{-2}A^{-1}] \cdot [L^2] = [ML^2T^{-2}A^{-1}]. \\$2. Self inductance $(L): \\$The formula for self inductance is $L = \frac{\Phi}{I},$ where $\Phi$ is magnetic flux and $I$ is current.$\\$Using the dimensions of magnetic flux from above, $[ML^2T^{-2}A^{-1}],$ and current $(I)$ with dimensions $[A], \\$the dimensions of self inductance are $\frac{[ML^2T^{-2}A^{-1}]}{[A]} = [ML^2T^{-2}A^{-2}]. \\$3. Magnetic permeability $(\mu): \\$The formula for magnetic permeability is $\mu = \frac{B}{H},$ where $B$ is magnetic field and $H$ is magnetic field strength.$\\$The dimensions of magnetic field strength $(H)$ are $[L^{-1}A]. \\$Thus, the dimensions of magnetic permeability are $\frac{[MT^{-2}A^{-1}]}{[L^{-1}A]} = [MLT^{-2}A^{-2}]. \\$4. Electric permittivity $(\varepsilon): \\$The formula for electric permittivity is $\varepsilon = \frac{D}{E},$ where $D$ is electric displacement and $E$ is electric field.$\\$The dimensions of electric displacement $(D)$ are $[L^{-2}AT],$ and electric field $(E)$ has dimensions $[MLT^{-3}A^{-1}]. \\$Thus, the dimensions of electric permittivity are $\frac{[L^{-2}AT]}{[MLT^{-3}A^{-1}]} = [M^{-1}L^{-3}T^4A^2]. \\$From the above analysis, the dimensions $[MLT^{-2}A^{-2}]$ correspond to magnetic permeability.$\\$Therefore, the correct option is 3: Magnetic permeability.