Electrostatic force between two identical charges placed in vacuum at distance of r is F . A slab of width r…

Force between the two charges when it is placed in vacuum and distance between them is $r$ is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$.

Now, the force when a dielectric $\left(k\right)$ is inserted between these two charges is $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{k{r}^{\text{'}2}}$.

On comparing the dielectric force with the vacuum force we will get, $r^2=k{r}^{\text{'}2}$

Or $r^{\text{'}}=r\sqrt{k}$

Hence, the effective thickness of slab is $\frac{r}{5}\sqrt{k}$.

Thus, the force between the charges is $F^{\text{'}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{{\left({\displaystyle \frac{r}{5}}+{\displaystyle \frac{r}{5}}\sqrt{9}\right)}^2}=\frac{q_1{q}_2}{4\pi {\epsilon }_0}\times \frac{25}{16r^2}=\frac{25}{16}F$