Answer
$2000^{\circ}C$.
Work Step by Step
We can calculate the resistance of the filament when the flashlight is on and the filament is hot.
$$R=\frac{V}{I}=\frac{3.0V}{0.20A}=15\Omega$$
Relate this resistance to the room temperature resistance to find the operating temperature.
$$R=R_0(1+\alpha(T-T_0))$$
$$T=T_0+\frac{1}{\alpha}(\frac{R}{R_0}-1)= 20^{\circ}C+\frac{1}{0.0045C^{-1}}(\frac{15\Omega}{1.5\Omega}-1)\approx 2000^{\circ}C $$