Option A $2\sqrt 2$
$AB = 6m , \ BC = 6m , \ CD = 2m, \ ED = 4m$
$\angle BAC = 60^{\circ}, \angle BCA = 60^{\circ}$
$so, \angle ABC = (180^{\circ}-60^{\circ}-60^{\circ}) = 60^{\circ}$
and $AC = 6m (\text{Equilateral triangle})$
$A$ is the starting point. Vivek walks $6$ meters North-East to point $B,$ then turns and walks $6$ meters South-East to point $C.$ He further moves $2$ meters South to the point $D$, $4$ meters West to point $E$, and $E$ is the endpoint.
As shown in figure, $AC=6m$ and $ED= FC = 4m.$
$\implies AF=AC- FC = 6-4=2m$
and $EF= CD = 2m$ and $\angle AFE = 90^{\circ}$
By Pythagoras theorem
$AE^2 = EF^2 + AF^2$
$\qquad = 4+4 = 8$
$\implies AE=2\sqrt 2.$