The `6Omega and 3 Omega` resistance are in parallel. Their equivalent resistance is `1/R=1/6+1/3 or R=2Omega`
Now, this `2Omega and 4Omega` resistanes are in series and their equivalent resistance is `4+2=6Omega`. Therefore, equivalent reistance of the network `=6 Omega`.
Current drawn from the battery is
`i=("netemf")/("net resistance")=18/6`
`=3A`