The given set M is $$\{-14,-11,-7,9,10,13\}$$ whose range $(= Highest value - Smallest value)$ is $13$$-(-14)=13+14=27$$

Median of the set $$M$$, having 6 values is the average of the $$3^{\text {rd \&$ the $4^{\text {th$$ term $$=\frac{-7+9}{2}=\frac{2}{2}=1 \&$$ the arithmetic mean (average) of the set $$M$$ is

$$=\frac{-14+(-11)+(-7)+9+10+13}{6}=\frac{0}{6}=0$$


Finally we can find the standard deviation of set M about the mean $$(=\overline{\mathrm{X)$$ i.e. about 0 so we get

$$\begin{tabular}{|l|l|l|} \hline x_i$ & x_i- Mean $=x_i-0 & \left(x_i-\text { Mean }\right)^2 \\ \hline-14 & -14 & 196 \\ \hline-11 & -11 & 121 \\ \hline-7 & -7 & 49 \\ \hline 9 & 9 & 81 \\ \hline 10 & 10 & 100 \\ \hline 13 & 13 & 169 \\ \hline & {$\sum\left(x_i-\bar{X}\right)^2=716$} \\ \hline \end{tabular}$$

So, the standard deviation is $$\sqrt{\frac{\sum\left(x_i-\bar{X}\right)^2}{n}}=\sqrt{\frac{716}{6}}=\sqrt{119.3} \approx 10.92$$

Now, checking from the statements

I. Median is greater than mean - which is true as Median $$=1>$$ Mean $$=0$$

II. Standard deviation is greater than range - which is false as standard deviation $$=10.92<$ Range $=27$$

III. Mean is greater than median - which is false as Mean $$=0$< Median $=1$$

Hence only statement I is true, so the answer is (A).