\(\begin{aligned}
& \left|\begin{array}{ll}
\log _5 729 & \log _3 5 \\
\log _5 27 & \log _9 25
\end{array}\right| \times\left|\begin{array}{ll}
\log _3 5 & \log _{27} 5 \\
\log _5 9 & \log _5 9
\end{array}\right| \\
& =\left|\begin{array}{ll}
6 \log _5 3 & \log _3 5 \\
3 \log _5 3 & \log _3 5
\end{array}\right| \times\left|\begin{array}{ll}
\log _3 5 & \frac{1}{3} \log _3 5 \\
2 \log _5 3 & 2 \log _5 3
\end{array}\right| \\
& =3 \log _5 3 \log _3 5\left|\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right| \times 2 \log _3 5 \log _5 3\left|\begin{array}{cc}
1 & 1 / 3 \\
1 & 1
\end{array}\right| \\
& =6[2-1] \times\left[1-\frac{1}{3}\right]=4
\end{aligned}\)
\(\because\) Hence the options
\(\log _3 5 \times \log _5 81=\left(\log _3 5\right) \times\left(4 \log _5 3\right)=4\)
Hence, option (d) is correct.