Given that page size is $256$ bytes $(2^8)$ and Main memory (MM) is 1KB $(2^{10}).$
So total number of pages that can be accommodated in MM $= \frac{2^{10}}{2^8} = 4.$
So, essentially, there are $4$ frames that can be used for paging (or page replacements).
The current sequence of pages in memory shows $3$ pages $(17, 1, 63).$ So, there is $1$ empty frame left. It also says that the least recently used page is $17.$
Now, since page size given is $8$ bits wide $(256\; B),$ and virtual memory is of $16$ bit, we can say that $8$ bits are used for offset. The given address sequence is hexadecimal can be divided accordingly:$$\small \begin{array}{|c|c|c|}\hline \textbf{Page Number} & \textbf{Offset} & \textbf{Page Number}\\
\textbf{in Hexadecimal} & \textbf{} & \textbf{in Decimal}
\\\hline \text{00} & \text{FF} & \text{0}\\ \text{01} & \text{0D} & \text{1} \\ \text{10} & \text{FF} & \text{16} \\ \text{11} & \text{B0} & \text{17} \\\hline \end{array}$$
We only need the Page numbers, which can be represented in decimal as: $0, 1, 16, 17.$
Now, if we apply LRU algorithm to the existing frame with these incoming pages, we get the following states:
$$\begin{array}{c|c}
0 & \text{Miss}&17\quad 1 \quad 63 \quad \bf{0}\\
1 & \text{Hit}&17\quad \textbf{1} \quad 63 \quad 0\\
16 & \text{Miss}&\textbf{16}\quad 1 \quad 63 \quad {0}\\
17 & \text{Miss}&16\quad 1 \quad \textbf{17} \quad {0}\\
\end{array}$$
- New status of the list is $\mathbf{16 \ 1 \ 17 \ 0}$.
- Number of page faults $\mathbf{= 3}$.
- Page replacements are indicated above.