Answer
The 6Ω, 3Ω are connected in parallel and the 2Ω resistor is connected in series to this
Work Step by Step
Power loss across resistors can be taken by the equation,
$$P=\frac{V^{2}}{R}$$
$9W= \frac{(6V)^{2}}{R}$
$R = 4Ω$
Since the three resistors provided are $6Ω$, $3Ω$ and $2Ω$ an equivalent resistance of $4Ω$ can be taken by having the $6Ω$ and $3Ω$ resistors in parallel and having $2Ω$ resistor in series to these
$R_{eq}=\frac{6Ω\times3Ω}{ (6 + 3)Ω} + 2Ω = 4Ω$
