For a non-pipelined system:
- Total number of instruction/task $(n)=500$
- Total time required to perform a single task in pipelined processor $(T_{np})=50$ ns
For a pipelined system:
- Total number of stages $(k)=6$
- Total number of instruction/task $(n)=500$
- Total time required to perform a single task in pipelined processor $(T_p)=10$ ns
$\because \text{Speedup ($S_k$)=$\frac{ET_{np}}{ET_p}$}$
$\implies S_k= \frac{(n*T_{np})}{(k+(n-1))T_p}$
$\implies S_k=\left[\frac{500*50}{(6+(500-1))*10}\right] $ ns
$\implies S_k=\left[\frac{25000}{(6+499)*10}\right]$ ns
$\implies S_k=\left[\frac{25000}{5050}\right]$ ns
$\implies S_k= 4.95$ ns
Option $(B)$ is correct.