The surface charge density $\sigma$ is defined as charge per unit area ($q/A$). For a sphere, the surface area is $A = 4\pi R^2$. Given: Diameter $d = 2.4$ m $\Rightarrow$ Radius $R = 1.2$ m Surface charge density $\sigma = 80.0 \, \mu\text{C/m}^2 = 80.0 \times 10^{-6}$ C/m$^2$

Calculation: $q = \sigma \times 4\pi R^2$ $q = 80.0 \times 10^{-6} \times 4 \times 3.1416 \times (1.2)^2$ $q = 80.0 \times 10^{-6} \times 12.566 \times 1.44$ $q \approx 1447.6 \times 10^{-6}$ C $q = 1.447 \times 10^{-3}$ C. (See NCERT Physics Class 12, Exercise 1.21).