For Pipelined system
There are five stages : S1 S2 S3 S4 S5 with time delays of 2ns, 2ns, 2ns, 3ns, 3ns. Also, the maximum buffer delay is 1ns.
Therefore, T for pipeline = (3+1)ns = 4ns.
Also, Efficiency = 0.8 so, $\frac{1}{S} = 0.8$ or S = $\frac{1}{0.8}$
So, total time taken for 100 instructions = $\left ( 100 + 5 - 1 \right )\frac{1}{0.8}4 = 520$ ns
For non-pipelined system
Total time taken = $\left ( 100 \times \left ( 2 + 2+ 2+ 3+ 3 \right ) \right ) = 1200$ ns
Speedup = $\frac{1200ns}{520ns} = 2.31$