Answer: 2140 bits.
Explanation:
In order to traverse the first network, the Data Link Layer builds a frame, which adds a 24-bit header to the IP datagram, which has 1820 bits (1500 bits of TCP data + 160 bits of TCP Header + 160 bits of IP Header).
When the frame arrives to the router that interconnects the two networks, the network layer is informed that the MTU for the destination network, is 800 bits only.
So, as the maximum size for a frame in the destination network must be 800 bits (including the 24-bit header), the IP protocol must fragment the original 1820-bit datagram, in a way that any frame have at a maximum 800 bits.
So the IP datagram is fragmented in 3 datagrams, two of them with 760 bits each (600 from the TCP segment + 160 bits from the IP Header), and a last one, with 620 bits (460 bits from the TCP segment + 160 bits from the IP Header).
All the datagrams are completed adding the 24-bit header to build frames that traverse the destination network until they reach to the final destination.
After stripping the 24-bit header, the frames (that became datagrams), arrive to the network layer protocol, divided in 3 fragments, one of 620 bits and two of 760 bits, with a total of 2140 bits, which, once reassembled, will become the original 1500 bits + 160 bits header of the TCP segment, after stripping the IP headers.
