$\quad F \propto \frac{1}{r^{3}}$

$\Rightarrow \text { Electric field, } E \propto \frac{1}{r^{3}} \Rightarrow E=\frac{k q}{r^{3}}$

Now, for Gaussian surface $(r < R)$.

Electric flux linked with surface is

$\phi=\int \frac{k q}{r^{3}} \cdot 2 \pi r d r$

$=2 \pi k q \int \frac{d r}{r^{2}}=2 \pi k q\left(-\frac{1}{r}\right)$

As flux is non-zero, charge density is also non-zero. Also, by symmetry of charge distribution electric field is zero.