(i) According to the question, the town is $10 \mathrm{~km}$ away, length of pair of copper wires used,
$$
\mathrm{L}=20 \mathrm{~km}=20000 \mathrm{~m}
$$
Resistance of $\mathrm{Cu}$ wires,
$$
\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\pi(\mathrm{r})^2}
$$
As given that, $\rho_{\mathrm{Cu}}=1.7 \times 10^{-8} ; \mathrm{r}=0.5 \mathrm{~cm}$
$$
=\frac{1.7 \times 10^{-8} \times 20000}{3.14\left(0.5 \times 10^{-2}\right)^2}=4 \Omega
$$
As given that, $\mathrm{P}=1 \mathrm{MW}=10^6$ watt
So, $\mathrm{I}$ at $220 \mathrm{~V}, \quad \mathrm{P}=\mathrm{VI}=10^6 \mathrm{~W}$,
$$
I=\frac{10^6}{220}=0.45 \times 10^4 \mathrm{~A}
$$
Power loss
$$
\begin{aligned}
\mathrm{P} &=\mathrm{RI}^2 \\
&=4 \times(0.45)^2 \times 10^8 \mathrm{~W} \\
&=8.26 \times 10^7 \mathrm{~W}
\end{aligned}
$$
Power loss in heating
$$
\begin{aligned}
&=82.6 \mathrm{MW} \\
\text { as } 82.6 \mathrm{MW} &>10^6 \mathrm{~W}
\end{aligned}
$$
So, this method cannot be used for transmission.
(ii) When power $\mathrm{P}=10^6 \mathrm{~W}$ is transmitted at $11000 \mathrm{~V}$.
$$
\mathrm{V}^{\prime} \mathrm{I}^{\prime}=10^6 \mathrm{~W}=11000 \mathrm{I}^{\prime}
$$
Current drawn, $\quad I^{\prime}=\frac{1}{1.1} \times 10^2$
$\therefore \quad \mathrm{R}_{\mathrm{Cu}}=4 \Omega$
(P) Power loss $=\mathrm{RI}^2=\frac{1}{1.21} \times 4 \times 10^4$
$\mathrm{P}=3.3 \times 10^4$ watt
So, Fraction of power loss $=\frac{3.3 \times 10^4}{10^6}=0.033$
Hence, $($ Power loss $)=3.3 \%$