(A.) Sequence number field is 32-bit, so 2$^{32}$ sequence number are possible. And 1 sequence number is assigned to 1-byte. Therefore 2$^{32}$ sequence number will be assigned to 2$^{32}$ Bytes. Hence max. value of L = 2$^{32}$ Bytes.
(B.) Given MSS = 536 Bytes, so number of segment = 2$^{32}/536$.
Given that 66 Byte of headers is added to each segment,
therefore, total amount of headers = (2$^{32}/536) * 66 Bytes$
total data to transmit = header + data = (2$^{32}/536) * 66 Bytes$ + 2$^{32}$ Bytes
Bandwidth = $155 Mbps$
Therefore, transmission time = total data * 8 bits / 155* 2$^{20}$ = 449 ms