Good catch on the missing series resister @Jerry-Hat-Trick, not a mistake you would expect from a "teacher".

 

R2 is a variable resistor assuming the LDR is 10.6k when dark. Otherwise you will have to improvise for your LDR dark resistance.
The LED you choose can be different from this blue LED specified as Forward voltage (Vf) = 3.45V at 20mA. measure the voltage across the LED.
The potentiometer is turned until LED turns on. In this example 20k * 53% is 10.6k ohm
Starting at 5 mA is safe that is R4 is 580 Ohm. I could later change R4 so that a superbright LED can draw 18mA. Which is about 53 Ohms,
I changed the source voltage to 8V. The variable resistor was again adjusted to check that the circuit will operate when the battery depletes.

 

Here, R2 represents the LDR.
In the original diagram provided by your teacher, the transistor base is connected to the middle terminal of the potentiometer. In this case, the potentiometer acts as a potential divider between the A and C terminals without any effect from the LDR. What you intend is to use the potentiometer as a variable resistor rather than a voltage divider. For this purpose, we use either the AB terminals or the BC terminals and leave one open.

 

​

a resistor to limit current through LED one and Q1 is not optional you must have a current living resistor or you risk smoking one or both Q1 and LED1.
LEDs, 555s, Flashers, and Light Chasers
Wendy's Index
A LDR (AKA CDC photoresistor) can go as low as >10Ω and as high as 1MΩ. If you are building this circuit I would strongly suggest you connect a DVM to it and measure its dark resistance and its dark resistance.