A physical quantity of the dimensions of length that can be formed out of $c$, $G$ and $\frac{e^2}{4\pi\varepsilon_0}$ is [$c$ is the velocity of light, $G$ is the universal constant of gravitation and $e$ is charge]

Let us write the dimensions of each quantity involved: Dimension of velocity of light, $c = [L T^{-1}]$ Dimension of universal gravitational constant, $G = [M^{-1} L^3 T^{-2}]$ From Coulomb's law, electrostatic force $F = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \implies \frac{e^2}{4\pi\varepsilon_0} = F r^2$. Therefore, the dimension of $\frac{e^2}{4\pi\varepsilon_0}$ is $[M L T^{-2}] [L^2] = [M L^3 T^{-2}]$. Let the physical quantity with dimension of length $L$ be related to $c$, $G$, and $\frac{e^2}{4\pi\varepsilon_0}$ as: $L \propto c^x G^y \left(\frac{e^2}{4\pi\varepsilon_0}\right)^z$ Substituting their dimensions: $[L^1] = [L T^{-1}]^x [M^{-1} L^3 T^{-2}]^y [M L^3 T^{-2}]^z$ $[M^0 L^1 T^0] = [M^{-y+z} L^{x+3y+3z} T^{-x-2y-2z}]$ Equating the powers of $M$, $L$, and $T$ on both sides, we get: For $M$: $-y + z = 0 \implies y = z$ For $T$: $-x - 2y - 2z = 0 \implies x + 4y = 0 \implies x = -4y$ For $L$: $x + 3y + 3z = 1 \implies -4y + 3y + 3y = 1 \implies 2y = 1 \implies y = 1/2$ Thus, $z = 1/2$ and $x = -4(1/2) = -2$. So, the physical quantity is $c^{-2} G^{1/2} \left(\frac{e^2}{4\pi\varepsilon_0}\right)^{1/2} = \frac{1}{c^2}\left[G\frac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$.