The displacement vector can be written as 
 $\overline{S}=20\widehat{i}+10\widehat{j}km$
Where  $\widehat{i}$, and $\widehat{j}$ are unit vectors along east and north (or) along X-axes and Y-axes respectively
$\therefore Magnitudeofdisplacement\left|S\right|=\sqrt{{20}^2+{10}^2}$

$\begin{array}{l}=\sqrt{500}\\ =22.36\mathrm{km}\end{array}$