A power station sends 360 kW to a town at 12 kV…

Step 1. The current. $I = \dfrac{P}{V} = \dfrac{360 \times 10^3\ \text{W}}{12 \times 10^3\ \text{V}} = 30$ A.

Step 2. The heat. $P_{loss} = I^2r = (30\ \text{A})^2 \times 8.0\ \Omega = 7200\ \text{W} = 7.2$ kW.

Step 3. The share. $\dfrac{P_{loss}}{P} = \dfrac{7.2\ \text{kW}}{360\ \text{kW}} = 0.020$, which is $2.0\ \%$.