The unit of permittivity of free space ε₀ is

The correct option is Coulomb²/Newton-metre².

According to Coulomb's Law of electrostatics, the force ($F$) between two point charges $q_1$ and $q_2$ separated by a distance $r$ in free space is given by the formula:

$F=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r^2}$

Here, ${\epsilon }_0$ is the permittivity of free space.

To find the unit of ${\epsilon }_0$, we can rearrange the formula to express ${\epsilon }_0$ in terms of the other variables:

${\epsilon }_0=\frac{q_1{q}_2}{4\pi F{r}^2}$

Now, let us substitute the Standard International (SI) units for each of the quantities:
- Charge ($q$) is measured in Coulomb (C).
- Force ($F$) is measured in Newton (N).
- Distance ($r$) is measured in metre (m).
- The constant $4\pi$ is dimensionless and has no unit.

Substituting these units into our expression, we obtain:

$\text{Unit of }{\epsilon }_0=\frac{\text{Coulomb}\times \text{Coulomb}}{\text{Newton}\times {\text{metre}}^2}$

Simplifying the unit block:

$\text{Unit of }{\epsilon }_0=\frac{{\text{Coulomb}}^2}{{\text{Newton-metre}}^2}$

Thus, the unit is Coulomb²/Newton-metre².