The correct option is B 3Ω
In this case, the 3 resistors of 1Ω on the upper half of the circuit are connected in series and the combined resistance R1 is given as 1+1+1=3Ω .
The 3 resistors of 2 ohms on the lower half of the circuit are connected in series and the combined resistance R2 is given as 2+2+2=6Ω.
The resistances R1 , R2 and 2Ω in the middle are connected in parallel. So the combined resistance R3 of this parallel connection is given as 13+12+16=11=1ohm. That is, the resistance in the closed loop is reduced to 1 ohm.
Now the total resistance R across the circuit is given as R1+R2+R3=1+1+1=3Ω.
Hence, the value of the resistance in the circuit from A to B is 3Ω.