Given: Mass of the particle $m=2\times {10}^{-6} \mathrm{kg}$, charge on the particle $q=5\times {10}^{-6} \mathrm{C}$.

Electric field due to conducting surface is given as $E=\frac{\sigma }{{\epsilon }_0}$, where $\sigma$ is the charge density and ${\epsilon }_0$ is the permittivity of free space.

The force exerted by the electric field will balance the weight of the charge.

$Eq=mg$

$\frac{\sigma }{{\epsilon }_0}q=mg$

$\Rightarrow \sigma =\frac{mg{\epsilon }_0}{q}$

$\frac{2\times {10}^{-6}\times 10\times 8.85\times {10}^{-12}}{5\times {10}^{-6}}=35.4\times {10}^{-12} \mathrm{C} {\mathrm{m}}^{-2}$