Let ${{[}}{{{\varepsilon }}_{{0}}}]$ denote the dimensional formula of the permittivity of the vacuum, and ${{[}}{{{\mu }}_{{0}}}{{]}}$ that of the permeability of the vacuum. If ${{M = }}$mass, ${{L = }}$length, ${{T = }}$time and ${{I = }}$electric current.
(This question has multiple correct options)
$
  {{(A) [}}{{{\mu }}_{{0}}}{{] = }}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{2}}}{{I}} \\
  {{(B) [}}{{{\varepsilon }}_{{0}}}{{] = }}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^4}{{{I}}^{{2}}} \\
  {{(C) [}}{{{\mu }}_{{0}}}{{] = ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}} \\
  {{(D) [}}{{{\mu }}_{{0}}}{{] = M}}{{{L}}^{{2}}}{{{T}}^{{{ - 1}}}}{{I}} $

Hint: For finding the dimensional formula of permittivity in vacuum, remind the formula in which this term appears. Then re-arranging accordingly and find out the dimensional formula for the same. And then follow the same for finding the dimensional formula of permeability of the vacuum.

Complete step by step solution:
Given: ${{M = }}$mass, ${{L = }}$length, ${{T = }}$time and ${{I = }}$electric current
${{[}}{{{\varepsilon }}_{{0}}}]$ denotes the dimensional formula of the permittivity of the vacuum
 ${{[}}{{{\mu }}_{{0}}}{{]}}$ denotes the dimensional formula of the permeability of the vacuum
Using coulomb’s law, we have
${{F = }}\dfrac{{{1}}}{{{{4\pi }}{{{\varepsilon }}_{{0}}}}}\dfrac{{{{{q}}_{{1}}}{{{q}}_{{2}}}}}{{{{{r}}^{{2}}}}}$
On rearranging the terms, we have
$\Rightarrow {{{\varepsilon }}_{{0}}}{{ = }}\dfrac{{{1}}}{{{{4\pi F}}}}\dfrac{{{{{q}}_{{1}}}{{{q}}_{{2}}}}}{{{{{r}}^{{2}}}}}$
Let us assume that both the charges are equal. Mathematically, we can say that $\Rightarrow {{{q}}_{{1}}}{{ = }}{{{q}}_{{2}}}{{ = q}}$.
Now, the formula becomes
$\Rightarrow {{{\varepsilon }}_{{0}}}{{ = }}\dfrac{{{1}}}{{{{4\pi F}}}}\dfrac{{{{{q}}^2}}}{{{{{r}}^{{2}}}}}$
The dimensional formula of charge (q) is ${{[IT]}}$
The dimensional formula of force (F) is ${{[ML}}{{{T}}^{{{ - 2}}}}{{]}}$
The dimensional formula of distance (r) is ${{[L]}}$
Now, the dimensional formula of permittivity of the vacuum ${{{\varepsilon }}_{{0}}}$ is:
$\Rightarrow {{{\varepsilon }}_{{0}}}{{ = }}\dfrac{{{{[}}{{{I}}^{{2}}}{{{T}}^{{2}}}{{]}}}}{{{{[ML}}{{{T}}^{{{ - 2}}}}{{{L}}^{{2}}}{{]}}}}{{ = [}}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{4}}}{{{I}}^{{2}}}{{]}}$
Or ${{[}}{{{\varepsilon }}_{{0}}}{{] = }}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{4}}}{{{I}}^{{2}}}$
Formula for speed of light is ${{c = }}\dfrac{{{1}}}{{\sqrt {{{{\varepsilon }}_{{0}}}{{{\mu }}_{{0}}}} }}$
On rearranging terms, we get
$\Rightarrow {{{\mu }}_{{0}}}{{ = }}\dfrac{{{1}}}{{\sqrt {{{{\varepsilon }}_{{0}}}{{{c}}^{{2}}}} }}$
The dimensional formula for permittivity of vacuum is ${{{\varepsilon }}_{{0}}}{{ = [}}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{4}}}{{{I}}^{{2}}}]$ (as calculated above)
The dimensional formula of speed of light ${{c = [L}}{{{T}}^{{{ - 1}}}}{{]}}$
Thus, the dimensional formula of permeability of the vacuum is given by
$\Rightarrow {{{\mu }}_{{0}}}{{ = }}\dfrac{{{1}}}{{{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{4}}}{{{I}}^{{2}}}{{{L}}^{{2}}}{{{T}}^{{{ - 2}}}}}}{{ = [ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}}{{]}}$
Or ${{[}}{{{\mu }}_{{0}}}{{] = ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}}$
Thus, the dimensional formula of permittivity of the vacuum is ${{[}}{{{\varepsilon }}_{{0}}}{{] = }}{{{M}}^{{{ - 1}}}}{{{L}}^{{{ - 3}}}}{{{T}}^{{4}}}{{{I}}^{{2}}}$ and the dimensional formula of permeability of the vacuum is ${{[}}{{{\mu }}_{{0}}}{{] = ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}}$.

Therefore, option (B) and (C) are the correct choices.

Note: Dimensional formula will always be closed in a square bracket [ ]. Here, in the options none of the dimensional formulas are kept in square brackets. This is so because the square brackets are imposed on the quantity here. For example, if we write the dimensional formula of permeability of vacuum, ${{{\mu }}_{{0}}}{{ = [ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}}]$ or ${{[}}{{{\mu }}_{{0}}}{{] = ML}}{{{T}}^{{{ - 2}}}}{{{I}}^{{{ - 2}}}}$. They both are the same thing.