$S.D. = \sqrt {\frac{{\sum {{{\left( {x – \bar x} \right)}^2}} }}{n}} $

$\bar x = \frac{{\sum x }}{4} = \frac{{ – 1 + 0 + 1 + k}}{4} = \frac{k}{4}$

Now $\sqrt 5 = \sqrt {\frac{{{{\left( { – 1 – \frac{k}{4}} \right)}^2} + {{\left( {0 – \frac{k}{4}} \right)}^2} + {{\left( {1 – \frac{k}{4}} \right)}^2} + {{\left( {k – \frac{k}{4}} \right)}^2}}}{4}} $

$ \Rightarrow 5 \times 4 = 2{\left( {1 + \frac{k}{{16}}} \right)^2} + \frac{{5{k^2}}}{8}$

$ \Rightarrow 18 = \frac{{3{k^2}}}{4}$

$ \Rightarrow {k^2} = 24$

$ \Rightarrow k = 2\sqrt 6 $