The point charges $3\mu \mathrm{C}\text{ and }4\mu \mathrm{C}$ are placed at a separation of 7 m.  The medium between them is of two types as shown in figure, the electric force acting between them is

For finding air equivalent separation

We have $\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^{\mathrm{\prime }2}}=\frac{1}{4\pi {\in }_0{\in }_{\mathrm{r}}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$      ${\mathrm{F}}^{\mathrm{\prime }}=\frac{1}{4\pi {\in }_0}\frac{{\mathrm{q}}_2{\mathrm{q}}_2}{{\mathrm{r}}^{\mathrm{\prime }2}}$

$\Rightarrow {\mathrm{r}}^{\mathrm{\prime }}=\sqrt{{ϵ}_{\mathrm{r}}}\mathrm{r}$       $=1.73\times {10}^{-4}\mathrm{N}$

$\therefore r_{\text{equivalent }}=\left(\sqrt{9}\right)3+\left(\sqrt{16}\right)4=25\mathrm{m}$

Therefore , F' =$\frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{q_1{q}_2}{{\left(r_{equivalent}\right)}^2}=9\times {10}^9\times \frac{3\times 4\times {10}^{-12}}{{25}^2} N =1.732\times {10}^{-4} N$