Answer
See the detailed answer below.
Work Step by Step
We first need to find the equivalent resistance that makes the battery give a power of 9 W.
We know that the power delivered by the battery is given by
$$P=\dfrac{(\Delta V_{\rm battery})^2}{R_{\rm eq}}$$
Hence,
$$R_{\rm eq}=\dfrac{(\Delta V_{\rm battery})^2}{P}=\dfrac{6^2}{9}=\bf 4\;\rm Omega$$
So we need to find a net resistance of 4 $\Omega$ from the 3 given resistors. We can get that by connecting the two resistors of 3 $\Omega$ and 6 $\Omega$ in parallel as one combination. This combination has a net resistance of 2 $\Omega$.
Then add this combination to the third resistor of 2 $\Omega$ in series. Then we get a net resistance of 4 $\Omega$.
See the figure below.
