Answer
$9.0\Omega$.
Work Step by Step
Find the current in the circuit from the bulb’s power rating.
$$P=IV$$
$$I=\frac{P}{V}=\frac{2.0W}{3.0V}=0.667A$$
Note that the bulb is to have a voltage of 3.0 V across it, so from Kirchhoff’s loop law, the resistor R has 6.0 V across it.
Find the resistance of R from Ohm’s Law.
$$R=\frac{V_R}{I_R}=\frac{6.0V}{0.667A}=9.0\Omega$$