A current repeats this pattern. It is +6.0 A for 3.0 ms…

Step 1. The mean. Add current times time for each part: $6.0\ \text{A} \times 3.0\ \text{ms} - 2.0\ \text{A} \times 1.0\ \text{ms} = 16\ \text{A ms}$. Divide by the period: $I_{mean} = \dfrac{16\ \text{A ms}}{4.0\ \text{ms}} = 4.00$ A.

Step 2. The mean square. Add the squares times the times: $(6.0\ \text{A})^2 \times 3.0\ \text{ms} + (2.0\ \text{A})^2 \times 1.0\ \text{ms} = 112\ \text{A}^2\text{ms}$. Divide by the period: $\dfrac{112\ \text{A}^2\text{ms}}{4.0\ \text{ms}} = 28.0\ \text{A}^2$.

Step 3. The rms current. $I_{rms} = \sqrt{28.0\ \text{A}^2} = 5.29$ A.

Step 4. The power. $P = I_{rms}^2 R = 28.0\ \text{A}^2 \times 5.0\ \Omega = 140$ W.