Not sure why the ON resistance of the LDR is an issue. For Vout to be at a high value the OFF resistance of the LDR must be high. If we assume the OFF resistance is infinity, then Vout is given by the voltage divider of the 4.3K bias resistor and the 1.5K base resistor. Vout = (5-0.6)(1.5/(1.5+4.3))+0.6 = 1.74V so Vout can never reach 3.2V no matter what LDR is used.
Sorry, I was assuming Vout was the voltage on the collector! (I couldn't read the diagram - it looked like omega 2t or something. I've been spending too much time on Bhuvanish's questions! And why is Vout on the input?)
BTW I still can't work out what Collin was hinting at.
Ok, so Vout isn't 3.2V but 1.74 V and can create about 0.75mA base current, which should energise the relay.
My worry was that the ON resistance was too high at say 2k, because it would not pull the base low enough to cut off the transistor. At 2k, the voltage divider of 4k3 and 2k gives Vout 1.58V ,if no base current, but that's enough to draw base current (I make it about 0.34mA , giving 35+mA collector current) That may not be enough to pull in the relay, but it may be too high to let it drop out.. (If it were a good BC109, Ic would be well over 100mA and the relay would be always on.)
Looking through Farnell I couldn't find an LDR lower than 2k (at 10 lux, however much that is, but it's what they quote On resistance at.)