Read the topology first. R₁ is in series with the source, and R₂, R₃, R₄ all hang between the same two nodes, so they are in parallel. Find the voltage across that parallel group and the rest follows.

Step 1 — combine the parallel resistors.

\(\dfrac{1}{R_p}=\dfrac{1}{33.3}+\dfrac{1}{50}+\dfrac{1}{20}=0.0300+0.0200+0.0500=0.1000\)

\(R_p=10\ \Omega\)

(The numbers are chosen to make this exact: 33.3 Ω contributes 0.03 S, so the three conductances add to a round 0.1 S.)

Step 2 — total resistance and source current.

\(R_T=R_1+R_p=40+10=50\ \Omega\)

\(I=\dfrac{100}{50}=2\ \text{A}\)

Step 3 — voltage across the parallel section. This is the voltage that appears across R₄:

\(V_p=I\,R_p=2\times10=20\ \text{V}\)

(Equivalently, the 40 Ω drops 80 V, leaving 20 V of the 100 V supply.)

Step 4 — power in R₄.

\(P_4=\dfrac{V_p^{2}}{R_4}=\dfrac{20^{2}}{20}=\dfrac{400}{20}\)

\(P_4=20\ \text{W}\)

Cross-check by current division. The current in R₄ is \(20/20=1\ \text{A}\), so \(P_4=I^{2}R_4=1^{2}\times20=20\ \text{W}\) ✓. As a further check, the three branch currents 0.6 A, 0.4 A and 1 A sum to the 2 A supplied ✓.

Where the distractors come from. 100 W is the total power drawn from the source (\(100\times2=200\) W, halved if you slip on the current); 10 W would follow from wrongly taking 10 V across the group, and 15 W from using the wrong branch.

Hence, the power dissipated in R₄ is 20 watts.