A kettle on a mains supply draws the power p(t) = 1800 sin²(100π t) watts, with t in seconds…

Step 1. The mean over one cycle. $\dfrac{1}{0.02}\displaystyle\int_0^{0.02} 1800\sin^2(100\pi t)\,dt$. Write $\sin^2 u = \dfrac{1 - \cos 2u}{2}$. The cosine term averages to $0$ over a whole number of cycles, so the mean of $\sin^2$ is $\dfrac{1}{2}$.

Step 2. The mean power. $1800 \times \dfrac{1}{2} = 900$ W, which is $0.9$ kW.

Step 3. The energy. $0.9\ \text{kW} \times 3\ \text{h} = 2.7$ kWh.

Step 4. The cost. $2.7\ \text{kWh} \times 0.30\ \text{euros per kWh} = 0.81$ euros.