A metal cube of side 0.1 m is uniformly charged by 18 μC. Its surface charge density will be

$3 × 10^{-4}\, Cm^{-2}$

The correct answer is Option (3) → $3 × 10^{-4}\, Cm^{-2}$

Surface charge density $\sigma$ is given by:

$\sigma = \frac{Q}{A}$

Where:
$Q = 18 \times 10^{-6} \ \text{C}$ (total charge)
$A = 6 \times (\text{side})^2 = 6 \times (0.1)^2 = 6 \times 0.01 = 0.06 \ \text{m}^2$ (total surface area of the cube)

So,

$\sigma = \frac{18 \times 10^{-6}}{0.06} = 3 \times 10^{-4} \ \text{C/m}^2$