From the data sheet, the buzzer will work at a minimum of about 4 volts. If the transistor works as a perfect switch, 8 volts will appear over the series combination of buzzer and R2. If 4 volts is the buzzer voltage, the voltage over R2 is also 4 volts, and for a 500 ohm resistor, that implies a current of 8 mA.
In order to get this much current through the load, the base current should be about 1/10 this, or 0.8 ma. However, the voltage across R1 will be (about) 7 volts, and 7 volts divided by 50k is 0.14 mA, which is too small by a factor of about 6.
To compensate, reduce R1 to about 5k. At the same time, reduce R2 to something like 100 ohms or less. For that matter, you can reduce R2 to about 30 - 50 ohms and get maximum loudness from the buzzer. If you do this, you'll need to supply about 30 mA to the buzzer, which will imply a value of R1 of about 230 ohms. There is a little bit of uncertainty here, since if R2 is zero you run the risk of putting more than the rated 7 volts on the buzzer. Frankly, I doubt a little more than 7 volts on the buzzer will damage it - but there are no guarantees when you exceed the manufacturer's maximum specifications.
When you do this, I'm afraid you'll find that the turn-on light level will change as well, and the buzzer will sound when it's not all that dark. It will depend on exactly what LDR you use.