Let `[epsilon_(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`,

Explanation

From coulomb's law `F = (1)/(4pi epsilon_(0)) (q_(1)q_(2))/(R^(2))`
`epsilon_(0) = (q_(1) q_(2))/(4pi FR^(2))`
Substituting the units
`epsilon_(0) = (c^(2))/(N-m^(2)) = ([AT]^(2))/([MLT^(-2)][L^(2)])`
`[M^(-1) L^(-3) T^(4) A^(2)]`