I drew up a circuit for a dark-activated motor here:
https://www.electronicspoint.com/question-mosfet-t258206.html#post1536817. The diode isn't needed because a light bulb is not an inductive load.
You can replace the motor with a light bulb. As duke37 says, you need to keep the LDR away from the light source otherwise the circuit will oscillate.
I recommend using a MOSFET with a low Vgs saturation voltage such as the Fairchild FDP8880 (see
http://www.digikey.com/product-detail/en/FDP8880/FDP8880FS-ND/976840). Whatever MOSFET you use, make sure it saturates with 5V gate-source voltage.
If you need to reduce the 5V down to 3V for the light bulb, you can use a series resistor. Assuming your bulb draws 200 mA at 3V (which I think is what you're saying in your first post), the resistor needs to drop 2V with 200 mA flowing through it, so according to Ohm's Law (R = V / I), its resistance needs to be 2 / 0.2 which is 10 ohms.
The resistor will dissipate some power. From the power law, P = V I, power dissipation will be 2 * 0.2 which is 0.4 watts. I would use a 1 watt resistor and mount it off the board. A larger resistor, such as 3W or 5W, will stay cooler because it has a larger surface area.