The dimensional formula of permeability of free space μ₀ is

The correct option is [MLT⁻²A⁻²].

To find the dimensional formula of the permeability of free space (${\mu }_0$), we can use the formula for the magnetic force per unit length between two parallel, current-carrying conductors separated by a distance $d$:
$\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$
where:
- $F$ is the force between the conductors.
- $L$ is the length of the conductors.
- $I_1$ and $I_2$ are the currents flowing through the conductors.
- $d$ is the separation distance between them.

Rearranging the formula to solve for ${\mu }_0$, we get:
${\mu }_0=\frac{2\pi dF}{L{I}_1{I}_2}$

Now, we substitute the dimensional formulas of the individual physical quantities into the equation. Note that the constant $2\pi$ is a dimensionless quantity.
- Dimension of force, $\left[F\right]=\left[ML{T}^{-2}\right]$
- Dimension of distance, $\left[d\right]=\left[L\right]$
- Dimension of length, $\left[L\right]=\left[L\right]$
- Dimension of electric current, $\left[I_1\right]=\left[I_2\right]=\left[A\right]$

Substituting these dimensions into the expression for ${\mu }_0$:
$\left[{\mu }_0\right]=\frac{\left[L\right]·\left[ML{T}^{-2}\right]}{\left[L\right]·[A}$

$]·[A]$

Simplifying the expression:
$\left[{\mu }_0\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L{A}^2\right]}$
$\left[{\mu }_0\right]=\left[ML{T}^{-2}{A}^{-2}\right]$

Thus, the dimensional formula of the permeability of free space ${\mu }_0$ is [MLT⁻²A⁻²].