$\underline{\textbf{Answer:}\Rightarrow}\;\;4.76$
$\underline{\textbf{Explanation:}\Rightarrow}$
Number of stalls $=5$
Clock Time $=20\;\text{ns}$
Number of instructions $=100$
Time taken by non-pipelined system $=5\times 20\times 100 = 10000$
Time taken by a pipelined system if no stall is present $=5\times \enclose{circle}{1}\times 20 + \enclose{circle}{99}\times 20 + \mathbf{19\;stalls} = 2099 \;\;\text{[Since, stall in the last cycle is insignificant.]}$
$\mathrm{5, stall, 10, stall, 15, stall, 20, stall, 25, stall, 30, stall, 35, stall, 40, stall, 45, stall, 50, stall, 55, stall, 60, stall, 65, stall, 70, stall, 75, stall, 80, stall, 85, stall, 90, stall, 95, stall, 100, \require{cancel} \cancel{stall}=19 \;stalls}$
$\therefore \;\text{Speedup} =\dfrac{\text{Time without pipelining}}{\text{Time with pipelining}}= \dfrac{10,000\;\text{ns}}{2099\;\text{ns}} = 4.76$