A physical quantity X is related to the permittivity of free space...

First, write the dimensional formulas for the given fundamental constants:
Permittivity of free space $[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
Planck's constant $[h] = [M L^2 T^{-1}]$
Speed of light $[c] = [L T^{-1}]$

Substitute the given exponents into the relation $X \propto \varepsilon_0^{-1/2} h^{1/2} c^{1/2}$:
$[X] = [M^{-1} L^{-3} T^4 A^2]^{-1/2} [M L^2 T^{-1}]^{1/2} [L T^{-1}]^{1/2}$

Simplify the powers for each dimension:
For $M$: $(-1)(-1/2) + (1)(1/2) + (0) = 1/2 + 1/2 = 1$
For $L$: $(-3)(-1/2) + (2)(1/2) + (1)(1/2) = 3/2 + 1 + 1/2 = 3$
For $T$: $(4)(-1/2) + (-1)(1/2) + (-1)(1/2) = -2 - 1/2 - 1/2 = -3$
For $A$: $(2)(-1/2) = -1$

The dimensional formula for $X$ is $[M L^3 T^{-3} A^{-1}]$.

Now, check the dimensions of the given options:
Electric field $E = \dfrac{F}{q} = \dfrac{[M L T^{-2}]}{[A T]} = [M L T^{-3} A^{-1}]$
Electric flux $\Phi_E = E \times \text{Area} = [M L T^{-3} A^{-1}] \times [L^2] = [M L^3 T^{-3} A^{-1}]$
Electric potential $V = \dfrac{W}{q} = \dfrac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$
Magnetic flux $\Phi_B = [M L^2 T^{-2} A^{-1}]$

Therefore, $X$ represents electric flux.

Answer: Electric flux