break the complicated network of resistors down into managable chunks that you know how to deal with by using the rules for combining resistances in series and in parallel to replace 2 or more resistors with one equivalent resistor.
the rule for series resistances is Requiv=R1+R2...
the rule for parallel resistances is 1/Requiv=(1/R1)+(1/R2)...
step 1: you have one resistor Z in series with the network consisting of WXY
if Z is 3 ohms then the resistance of the network WXY must be 2 ohms
step 2: combined resistance of the series resistors W and X in the upper branch = 3+3 ohms = 6 ohms
step 3: combined resistance of the resistor Y in parallel with the equivalent value for W and X... which must be equal to 2 (step 1)
1/Requiv=(1/R1)+(1/R2)...
1/2=(1/3) + (1/6)
1/2=(2/6) + (1/6)
1/2=3/6 which is true
might be helpful to redraw the circuit at each stage of simplification until you get the hang of it.