A physical quantity of the dimensions of length that can be formed out of c, G and $\frac{e^2}{4\pi {\epsilon }_0}$c  is velocity of light, G is universal constant of gravitation and e is charge]

Length in terms of velocity of light, gravitational constant, and  $\frac{e^2}{4\pi {ϵ}_0}:$

We know,  $\frac{e^2}{4\pi {ϵ}_0{r}^2}=F$

(from Coulomb’s law)

$\left[\frac{{\mathrm{e}}^2}{4\pi {ϵ}_0}\right]=\left[{\mathrm{Fr}}^2\right]=\left[{\mathrm{ML}}^3{\mathrm{T}}^{-2}\right]$

Gravitational constant

$\left[G\right]=\left[\frac{{\mathrm{Fr}}^2}{{\mathrm{m}}^2}\right]=\frac{{\mathrm{MLT}}^{-2}{\mathrm{L}}^2}{{\mathrm{M}}^2}={\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}$

From Newton’s law, $\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$

$\text{[Velocity of light]}={\mathrm{LT}}^{-1}$

Let  $L={\left(\frac{e^2}{4\pi {ϵ}_0}\right)}^x\left(G{)}^y\right(e{)}^z$

$\begin{array}{rl} & \text{Let}L={\left(\frac{e^2}{4\pi {ϵ}_0}\right)}^x\left(G{)}^y\right(e{)}^z\\ & L={\left(M{L}^3{T}^{-2}\right)}^x{\left(M^{-1}{L}^3{T}^{-2}\right)}^y(LT\\ & M^0{L}^1{T}^0=M^{x-y}{L}^{3x+3y+z}\cdot T^{-2x-2y}{y}^2\\ & x-y=0\to \left(1\right)x=y\\ & 3x+3y+z=1\to \left(2\right)\\ & -2x-2y-z=0\to \left(3\right)\end{array}$

Solving  $x=y=1/2$

Hence,  $L=\frac{1}{c^2}{\left(\frac{e^2}{4\pi {ϵ}_0}G\right)}^{1/2}$