You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors. I'm curious to know how you did that.
The three resistors at the left end of the diagram are 3Ω , 2Ω , and 1Ω all in series. They behave like a single resistor of (3+2+1) = 6Ω .
That 6Ω resistor is in parallel with the 2Ω drawn vertically in the middle of the diagram. That combination acts like a single resistor of 1.5Ω in that position.
Finally, we have that 1.5Ω resistor in series with 1Ω and 4Ω . That series combination behaves like a single resistor of 6.5Ω across the battery V.
