To solve this problem we must find the values of the equivalent resistances in both section 1 and section 2. Later we will calculate the total current and the total voltage. With the established values we can find the values of the currents in the 3 Ohms resistance and the power there.
The equivalent resistance in section 1 would be
[tex]R_{eq1} = \frac{(3\Omega)(6\Omega)}{(3+6)\Omega}[/tex]
[tex]R_{eq1} = 2\Omega[/tex]
The equivalent resistance in section 2 would be
[tex]R_{eq2} = R_{eq1} +4\Omega[/tex]
[tex]R_{eq2} = 6\Omega[/tex]
Now the total current will be,
[tex]I_t = \frac{V_t}{R_{eq2}}[/tex]
[tex]I_t = \frac{12V}{6\Omega}[/tex]
[tex]I_t = 2.0A[/tex]
Finally the total Voltage will be,
[tex]V = IR_{eq1}[/tex]
[tex]V = (2.0A)(2.0\Omega)[/tex]
[tex]V = 4V[/tex]
Since the voltage across the 3 and 6 Ohms resistor is the same, because they are in parallel, the current in section 3 would be
[tex]I_{3.0\Omega} = \frac{V}{R}[/tex]
[tex]I_{3.0\Omega} = \frac{4.0V}{3.0\Omega}[/tex]
[tex]I_{3.0\Omega} = 1.3A[/tex]
Finally the power ratio is the product between the current and the voltage then,
[tex]P_{3.0\Omega} = I_{3.0\Omega} V[/tex]
[tex]P_{3.0\Omega} = (1.3A)(4.0V)[/tex]
[tex]P_{3.0\Omega} = 5.3W[/tex]
Therefore the correct answer is D.
