A charge is distributed in space such that the charge density $\rho$ varies with position $x,\text{ as }\rho \propto x^2$.  At points close to the x-axis, the electric field has an x component only. If the magnitude E of the electric field varies with position $x,\text{ as }E\propto x^n$ , then n equals:

Consider the closed box shown in the figure. Two surfaces of the box, each having area A, are parallel to the yz plane and located at positions x and x + dx. Let E and E + dE be the electric fields at positions x and x + dx directed along the +ve x-axis. The volume of this box is Adx. The charge inside the box is $Q_{en}=\rho \left(x\right)Adx=\left(c{x}^2\right)\left(Adx\right)$  where c is a constant of proportionality. The electric flux cutting the box is  $\phi =(E+dE)A-EA=AdE$

Now using Gauss's theorem, we have

${\epsilon }_0\varphi =Q_{en}$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\epsilon }_{o}AdE=\left(c{x}^2\right)\left(Adx\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}dE=\frac{c}{{\epsilon }_0}{x}^{2}dx\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}E=\frac{c}{{\epsilon }_0}\int x^{2}dx=\frac{c}{3{\epsilon }_0}{x}^3+D\end{array}$

Since charge distribution is symmetrical about x = 0, the electric field must also be symmetrical about x = 0. Hence, the arbitrary constant of integration D must be zero. Hence ,$E\propto x^3$