Given, $r=1.2 \mathrm{~m}$,
$$
\sigma=80.0 \mu \mathrm{Cm}^{-2}=80 \times 10^{-6} \mathrm{Cm}^{-2} \text {, }
$$
$\varepsilon_0=8.854 \times 10^{12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ (taken)
(a) $q=$ ?
(b) $\phi=$ ?
(a) By relation,
$$
\begin{aligned}
\mathrm{q} &=\sigma \mathrm{A}=\sigma 4 \pi \mathrm{r}^2 \\
&=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^2 \\
&=8 \times 12.56 \times 1.44 \times 10^{-5} \\
&=144.7 \times 10^{-5}=1.447 \times 10^{-3} \mathrm{C}
\end{aligned}
$$
(b) For outside space, the charge on the conducting sphere behaves as a point charge situated at the centre of the sphere.
By Gauss's theorem, $\phi=\mathrm{q} / \varepsilon_0$
$$
=\frac{1.44 \times 10^{-3}}{8.854 \times 10^{-12}}=1.63 \times 10^8 \mathrm{Nm}^2 / \mathrm{C}
$$