Given:
Message size $= 2400$ bits.
The header of the transport layer $= 150$ bits.
Segment size at transport layer $=$ (Header + Message size) $= 150 + 2400 = 2550$ bits.
Maximum, transmission unit at destination network $= 900$ bits
So, At a time, data supported by the destination network $= 900 \ – 26 = 874$ bits.
Hence, $2550$ bits are divided into packets having a maximum of $874$ bits.
But $874$ is not divisible by $8,$
So:
Length of packet $1 = 872$ bits.
Length of packet $2 = 872$ bits.
Length of packet $3 = 806$ bits.
Total $= 2550$ bits.
Network layer packets are transmitted via two networks, each of which uses a $26$ – bit header. So, The number of bits, including headers delivered to the destination network is:
Packet $1$ (Data + Header) size $= 26 + 26 + 872 = 924$ bits.
Packet $2$ (Data + Header) size $= 26 + 26 + 872 = 924$ bits.
Packet $3$ (Data + Header) size $= 26 + 26 + 806 = 858$ bits.
Total $= 2706$ bits.