If $$\mu_0$$ and $$\varepsilon_0$$ are the permeability and permittivity of free space, respectively, then the dimension of $$\left(\frac{1}{\mu_0 \varepsilon_0}\right)$$ is :

The permeability of free space is denoted by $$\mu_0$$ and the permittivity of free space by $$\varepsilon_0$$.

Electromagnetic‐wave theory gives the relation between the speed of light $$c$$ and these two constants:
$$c \;=\;\frac{1}{\sqrt{\mu_0\,\varepsilon_0}} \quad -(1)$$

Square both sides of $$(1)$$ to obtain
$$c^{2} \;=\;\frac{1}{\mu_0\,\varepsilon_0} \quad -(2)$$

The dimensions of speed are $$[c] = L\,T^{-1}$$. Squaring this,
$$[c^{2}] = (L\,T^{-1})^{2} = L^{2}\,T^{-2} \quad -(3)$$

From $$(2)$$, $$\dfrac{1}{\mu_0\,\varepsilon_0}$$ has the same dimensions as $$c^{2}$$. Therefore,
$$\left[\frac{1}{\mu_0\,\varepsilon_0}\right] = L^{2}\,T^{-2}$$

Hence, the correct option is Option B $$\big(L^{2}\,T^{-2}\big)$$.