The speed up ratio = non -piped line time implementation for a task/pipedline time implementation for the same task
In this case speed up = 50 ns/10 ns=5
further for 100 task----in this case of non-piped line processor takes 50*100 ns= 5000 ns
for the piped line processor with 6 stage ,task= 100, time /task=10 ns,the total time is taken (100+6-1)*10=1050 ns.
hence speed up for 100 task = 5000 ns/1050 ns=4.76
since theritically maximum speed up is equal to nos of stages which is 6 in this which is never achieveable, hence max speed up in this case is 5 as above. hence the option (B) is correct.