We can find the standard deviation of $\(3,7,9,13 \& 18\)$ about the mean $\((=\overline{\mathrm{X)\)$ i.e. $\(\frac{3+7+9+13+18}{5}=\frac{50}{5}=10

(\)$\(\frac{ Arithmetic Mean}{Average} =\frac{ Sum of terms }{ Number of terms })\)$ so we get

\begin{tabular}{|l|l|l|}
\hline $x_i$ & $x_i-$ Mean $=x_i-10$ & $\left(x_i-\text { median }\right)^2$ \\
\hline 3 & -7 & 49 \\
\hline 7 & -3 & 9 \\
\hline 9 & -1 & 1 \\
\hline 13 & 3 & 9 \\
\hline 18 & 8 & 64 \\
\hline & {$\sum\left(x_i-\bar{x}\right)^2=132$} \\
\hline
\end{tabular}

So, the standard deviation is $\(\sqrt{\frac{\sum(x_i-\bar{X})^2}{n}\)\(=\sqrt{\frac{132}{5}}=\sqrt{26.2} \approx 5.12\)$

Hence the answer is (D).