This is the circuit you should get. If phototransistor Q2 does not draw current, then Q1's base gets current through R2. The base is at 0.7 V (always 0.7 V higher than the emitter), and the power supply is 3 V, so there's 3 V - 0.7 V = 2.3 V across R2. Then because of Ohm's Law the current through R2 = 2.3 V / 1 kΩ = 2.3 mA. Transistor Q1 will want to increase that 100-fold to get 230 mA collector current. R1 will limit that. If Q1 is on then the LED's cathode will be at around 0 V, and the anode 2 V higher, at 2 V (that's typical for a red LED). So there remains 1 V for resistor R1, and if we want 20 mA through it (and the LED) we apply Ohm's Law again: R = 1 V / 20 mA = 50 Ω. So R1 will make sure that the LED current won't go higher than 20 mA. To help with the breadboard I just threw together the little circuit shown in your question. I only had an IR phototransistor but it doesn't matter much for this, it still works the same. Anyway here are a couple of pictures of it working, hopefully you can see how the connections go:

Its non-inverting input pin#3 is clamped with a fix reference derived from the junction of the resistive divider formed by R2/R3.As this happens, the gate input goes high, the output consequently becomes low making the outputs of the buffer gates high. The result is the triggering of the transistor and the relay assembly. The connected load over the relay now flips into the intended actions.

To make the circuit extermely compact one button battery type is preferred here, quite akin to those used in calculators, watches, etc. In other words the circuit can be used like aday activated automatic switch or a darkness activated automatic switch, depending upon the user preference or the specific application need. This variable resistor is used for setting the triggering point of the gate when the light falling over the LDR reaches the desired specified intensity. Can be used to switch a lighting circuit 'ON' and 'OFF' based on light conditions (nightfall, daybreak) Resistor R5 introduces some level of hysteresisto the op amp response so that the output of the op amp does behave erratically during the twilight or the transition periods where the light level on the LDR is at the threshold points.The phototransistor base is floating, and I swapped the 1k for a 22k in my circuit to bias it correctly (I arrived at this value roughly, see below) and used a BC337 npn. Since the BC337 has lots of gain the 22k works well for the base current. EDIT - the breadboard circuit you have added looks correct (though it's hard to read..) so go ahead and try it. If it doesn't work let us know. Maybe change the resistor to 2k or larger if you are worried about blowing the LED. With the BJT turned OFF, the SCRand the load are also turned OFF in the presence of day light on the LDR. The circuit is now transformed into a darkness activated switch for the load or the connected lamp.