How to use Lightsensor relay with 3v output
Assumptions:


You have a 3V and 5V available from somewhere.
You stated that you want a relay. I trust you either need galvanic isolation or a lot of load current. Otherwise I would consider a solid state switch.
Your light sensor is a photodiode.


The Schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Explanation:


R1 and D2 convert the light into a voltage indication. You will need to select R1 based on the current of D2.
R2 and R3 are a voltage divider. You will need to determine (likely experimentally) the voltage at which you consider light to be detected. Then select R2 and R3 such that 5*R3/(R2 + R3) give the desired voltage and R2 + R3 is around 100k. C2 is there to filter this reference.
CMP1 is an anolog comparator. It's output is digital '1' when the + is greater than the - and digital '0' otherwise. Therefore when the light is below a certain level the voltage on the - input will fall below the reference and the comparator will output a '1' (in this case 5V).
R4, Q1, and D1 are to drive the relay coil. R4 determines the base current of Q1 which is operated in saturation or cutoff to turn the relay on or off respectively. D1 is needed to prevent damage to Q1 during coil discharge events.
RLY1 connects the load to the desired 3V supply.