Answer :
a)Diameter of the sphere $=2.4 \;m$
$r= 1.2 \;m$
Surface charge density $\sigma= 80 \;\mu c/m^2$
$\qquad= 80 \times 10^{-6}c/m^2$
Surface change density $=\large\frac{change}{surface area}$
Surface area $= 4 \pi r^2$
$\sigma = \large\frac{q}{4 \pi r^2}$
$q= \sigma \times 4 \pi r^2$
$\qquad= 80 \times 10^{-6} \times 4 \times 3.14 \times 1.2 \times 1.2$
$q= 1.4 10^{-3}\;C$
b) Total flux $=\phi$
Total flux leaving the surface $\phi = \large\frac{Total change}{\in_0}$
$\phi =\large\frac{q}{\in_0} =\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}}$
$\phi = 1.6 \times 10^8 N-m^2 /C$
Hence the flux leaving the surface of the sphere is $1.6 \times 10^8 N-m^2/C$