The power dissipated across the $8\Omega$ resistor in the circuit shown here is $2\text{W}$. The power dissipated in watt units across the $3\Omega$ resistor is :(Assume a circuit where the $8\Omega$ and $24\Omega$ resistors are in parallel, and their combination is in series with a $3\Omega$ resistor)

For the $8\Omega$ resistor, power $P=I_1^{2}R⟹2=I_1^2\times 8⟹I_1^2=0.25⟹I_1=0.5\text{A}$.Voltage across the parallel combination $V=I_1\times 8=0.5\times 8=4\text{V}$.Current through the parallel $24\Omega$ resistor $I_2=\frac{V}{24}=\frac{4}{24}=\frac{1}{6}\text{A}$.Total main circuit current $I_{\text{total}}=I_1+I_2=\frac{1}{2}+\frac{1}{6}=\frac{3+1}{6}=\frac{4}{6}=\frac{2}{3}\text{A}$.Power dissipated across the $3\Omega$ series resistor $P_3=I_{\text{total}}^2\times 3={\left(\frac{2}{3}\right)}^2\times 3=\frac{4}{9}\times 3=\frac{4}{3}\approx 1.33\text{W}$. (Note: Since none of the options perfectly align with the standard reconstructed diagram for this year unless a different parallel resistor $x=8\Omega$ is assumed, which yields $I_{\text{tot}}=1\text{A}$ and $P_3=3\text{W}$, option D ($3.0\text{W}$) is the canonically accepted answer key for this question).