To solve this problem, we need to find a physical quantity with the dimensions of length using $c, G,$ and $\frac{e^2}{4\pi\varepsilon_0}. \\$First, let's determine the dimensions of each quantity:$\\$• $c$ (velocity of light) has dimensions $[LT^{-1}]. \\$• $G$ (universal gravitational constant) has dimensions $[M^{-1}L^3T^{-2}]. \\$• $\frac{e^2}{4\pi\varepsilon_0}$ (Coulomb's constant times charge squared) has dimensions $[ML^3T^{-2}]. \\$We need to form a quantity with dimensions of length $[L]. \\$Assume the quantity is of the form $c^a G^b \left(\frac{e^2}{4\pi\varepsilon_0}\right)^c. \\$The dimensions of this expression are:$\\ [L^a T^{-a}] [M^{-b} L^{3b} T^{-2b}] [M^c L^{3c} T^{-2c}]. \\$Combine the dimensions:$\\ [M^{-b+c} L^{a+3b+3c} T^{-a-2b-2c}]. \\$For this to have dimensions of length $[L],$ we need:$\\ 1. -b + c = 0 \quad ($no mass dimension$) \\ 2. a + 3b + 3c = 1 \quad ($length dimension$) \\ 3. -a - 2b - 2c = 0 \quad ($time dimension$) \\$From equation 1: $c = b. \\$Substitute $c = b$ into equations 2 and 3:$\\ a + 3b + 3b = 1 \\ a + 6b = 1 \quad$(equation 2)$\\ -a - 2b - 2b = 0 \\ -a - 4b = 0 \quad$(equation 3)$\\$From equation 3: $a = -4b. \\$Substitute $a = -4b$ into equation 2:$\\ -4b + 6b = 1 \\ 2b = 1 \\ b = \frac{1}{2}. \\$Then $c = b = \frac{1}{2}. \\$Substitute $b = \frac{1}{2}$ into $a = -4b: \\ a = -4 \times \frac{1}{2} = -2. \\$Thus, the expression is: $c^{-2} G^{1/2} \left(\frac{e^2}{4\pi\varepsilon_0}\right)^{1/2}. \\$This corresponds to Option 4: $\frac{1}{c^2} \left[ G \frac{e^2}{4\pi\varepsilon_0} \right]^{1/2}. \\$Therefore, the correct option is Option 4.