A better way to solve would be
For non pipeline implementation, time to complete each instruction t = 5+6+11+8 = 30ns
For pipeline implementation, max duration for each stage T = 11+1 =12ns
For k instructions,
Total cycles for non-pipeline implementation = k*t
Total cycles for non-pipeline implementation = (k + n - 1)*T, where n = #stages
Consider 100 instructions
i) for non pipeline: C1 = 100 * 30 = 3000ns
ii) for pipeline: C2 = (100 + 4 -1)*12 = 103*12 = 1236ns
iii) Speedup = C1/C2 = 3000/1236 = 2.5 (approx)
Same holds true for 1000 instructions,
Hence answer is B