Here router has two incoming/outgoing hardware that is full-duplex which means we can send and receive at the same time.
Transmission time is $6.72$ (How to get this Transmission time = Packet size / Bandwidth)
Transmission time is time for a packet to get out/in of wire. Also, assume even though the router has two incoming/outgoing hardware but it has a single processing unit.
Now assume that at $t=0$ router is free. At $t=6.72$ two packets, $p_{1}\;\&\; p_{2}$ arrived from two incoming/outgoing hardware. Now next slot of two packets will arrive at $6.72+6.72$ before that processing of these two packets must be completed so that there will not be any kind of delay. Hence we just have $6.72$ maximum times to process each packet, hence for each packet, we can devote a maximum of $3.36.$
Hence the answer is $3.36.$
Now for further explanation, I will show when the first two packets are processed. At $t=0$ both packets start coming out of wire, at $t=6.72$ both packets completely coming out of wire, and we start the processing of one of them (let's say $p_{1}$) and next packets $p_{3}, p_{4}$ starts coming out.
At $t=6.72+3.36$ processing of $p_{1}$ done, start transmitting it through port $1$.
At $t=6.72+ 6.72$ processing of $p_{2}$ done start transmitting it through port $2$.
At this time next packets $p_{3} \;\&\; p_{4}$ arrived, and we start processing one of them (let's say $p_{3}$), at t=6.72+6.72+3.36 p1 completely transmitted., processing of p3 done, start transmitting it through port $1\;($as port $2$ is busy in transmitting packet $2)$
At $t=6.72+6.72+6.72$ transmitting of $p_{2}$ done. processing of $p_{4}$ done starts transmitting it through port $2,$ next packets $p_{5} \;\&\; p_{6}$ arrived.
.
.
.
continue