Total page faults $=6.$

$\small\begin{array}{|c|c|c|c|} \hline \quad\bf{4}\quad& \quad\bf{7} \quad& \quad \bf{6}\quad & \quad\bf{1}\quad&\quad\bf{7}\quad&\quad \bf{6}\quad &\quad \bf{1}\quad &\quad \bf{2}\quad &\quad \bf{7}\quad &\quad \bf{2}\quad \\ \hline & & \underset{\boxed{F}}6&6&6&6&6&6&\underset{\boxed{F}}7&7\\ \hline \hline &\underset{\boxed{F}}7 &7 &7&7&7&7&\underset{\boxed{F}}2&2&2\\ \hline

\hline \underset{\boxed{F}}{4}&4 & 4&\underset{\boxed{F}}1&1&1&1&1&1&1\\ \hline

\end{array} \implies 6\text{ faults}$

Another way of answering the same.

$\require{cancel} \begin{array}{|c|}\hline

\quad6\quad \\\hline 7\\\hline\quad\cancel {4} \quad 1\\ \hline

\end{array}$ $\require{cancel} \begin{array}{|c|}\hline

\quad6\quad \\\hline \cancel{7}{2}\\\hline 1\\ \hline

\end{array}$ $\require{cancel} \begin{array}{|c|}\hline

\quad\cancel {6}{7}\quad \\\hline{2}\\\hline1\\ \hline

\end{array}$$\require{cancel} \begin{array}{|c|}\hline

\quad7\quad \\\hline {2}\\\hline 1\\ \hline

\end{array}$ $\implies 3 \text{ faults}+3 \text{ initial access faults} = 6 \text{ page faults}$

**OR**

$\require{cancel} \begin{array}{|c|}\hline

\quad\cancel6 7\quad \\\hline\cancel {7}2\\\hline\cancel 4 1\\ \hline

\end{array} \implies 3 \text{ faults}+3 \text{ initial access faults} = 6 \text{ page faults}$