Force in air at a distance $r,$

$F= \frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$ ......(i)

Let $r’$ be the distance in the medium, therefore

$4F= \frac{q_1{q}_2}{4\pi {ϵ}_0\left(16\right)(r’{)}^2}$ ......(ii)

From (i) $÷$ (ii)

$\frac{1}{4} = \frac{16(r’{)}^2}{r^2}$

$\therefore r’=\frac{r}{8}$