wirespeed forwarding
To operate at "wirespeed" means to operate at the bandwidth of the physical wire (or medium/link). This would require for other delays (processing, queuing etc) to not degrade the performance so much that we can't practically match the "wire speed"
To take an example, say we have a physical wire that goes from A to B, and the bandwidth of the wire is 100Mbps. But we practically observe that data is being transferred only at 80 Mbps.
This would mean that we're not functioning at wirespeed — the reason could be anything — receiver taking too much time to accept, or sender is taking too long to send, or packets are coming in so fast that the receiver's buffer overflowed.
Now here, we have a Sender and a Receiver (the router is the receiver)
Sender's $T_t$ is $6.72 \mu s$ from one link. There are two such links from Sender to Receiver, hence sender can transmit two packets to the receiver in $6.72 \mu s$. ($T_p$ is not given, so we assume it is negligible)
Now, Receiver has received 2 packets in $6.72 \mu s$. Two more packets would arrive after additional $6.72 \mu s$ and so on.
So, to clear the two packets out, Receiver has got $6.72 \mu s$ maximum. Why? Because if Receiver forwards them in more time, then we are piling up packets in it's buffer, and achieving wirespeed wouldn't be possible.
Hence, per packet Receiver can take maximum $\frac{6.72 \mu s }{2}=3.36\mu s$.
Option B
If the receiver takes no time i.e. $0 \mu s$ then it's great. But we can allow the receiver to take a maximum of $3.36 \mu s$ to process each packet — and that's what the question asks for.