Although Akash's answer is correct , adding my answer just to elaborate bit more
Here router has two incoming/outgoing hardware which are full duplex which means we can send and receive at same time.
Transmission time is 6.72(How to get this Transmission time = Packet size / Bandwidth)
Transmission time is time for packet to get out/in of wire.
Also assume even though router has two incoming/outgoing hardware but it has single processing unit.
Now assume that at t=0 router is free.
at t=6.72 two packets p1 & p2 arrived from two incoming/outgoing hardware.
Now next slot of two packets will arrive at 6.72+6.72 before that processing of this two packets must be completed so that there will not be any kind of delay.
hence we just have 6.72 maximum time to process each packet , hence for each packet we can devote maximum 3.36.
Hence answer is 3.36.
Now for further explanation I will show when first two packets are processed.
at t=0 both packets starts coming out of wire
at t=6.72 both packets completely coming out of wire , and we start processing of one of them (lets say p1)
and next packets p3 , p4 starts coming out.
at t=6.72+3.36 processing of p1 done ,start transmitting it through port 1
at t=6.72+ 6.72 processing of p2 done start transmitting it through port 2.
at this time next packets p3 & p4 arrived , and we starts processing one of them (lets say p3)
at t=6.72+6.72+3.36 p1 completely transmitted., processing of p3 done , start transmitting it though port 1 (as port 2 is busy in transmitting packet 2)
at t=6.72+6.72+6.72 transmitting of p2 done. processing of p4 done starts transmitting it through port 2 , next packets p5 & p6 arrived