$\underline{\textbf{Answer:}\Rightarrow}$
$\underline{\textbf{Explanation:}\Rightarrow}$
Assume $\mathbf k$ instructions:
$\mathbf{k-1}$ instructions execution without page fault $ = 1 \times (\mathrm k-1) \;\text{nsec}$
$1$ instruction with page fault $ = (1+n) \times 1\;\text{nsec}$
$\therefore $ Average is given by the formula $ = {(1\times (\mathrm k-1) + 10 +n) \times 1)}{k}\; \mathbf{ =1 + \dfrac{n}{k} \;\text{nsec}}$