Let the man start from origin. He walks west side 4 metrs.

$\therefore\:\overrightarrow {OA}=-4\hat i$

Then he walks 3 metrs. $\large\frac{\pi}{6}$ east of north and moves to $B$

$\therefore \: |\overrightarrow {AB}|=3$, $< BAO=\large\frac{\pi}{3}$ and $ |\overrightarrow {OA}|=4$

Let $\overrightarrow {OB}=x\hat i+y\hat j$

From triangular law of addition $\overrightarrow {OA}+\overrightarrow {AB}=\overrightarrow {OB}$

$\Rightarrow\:-4\hat i+\overrightarrow {AB}=x\hat i+y\hat k$

$\Rightarrow\:\overrightarrow {AB}=x\hat i+y\hat k+4\hat i=(x+4)\hat i+y\hat k$

Also $cos\large\frac{\pi}{3}$$=\large\frac{\overrightarrow {OA}.\overrightarrow {AB}}{|\overrightarrow {OA}||\overrightarrow {AB}|}=\frac{(-4\hat i).((x+4)\hat i+y\hat j)}{12}$

$\Rightarrow\:\large\frac{1}{2}=\frac{-4x-16}{12}$

$\Rightarrow\:x=-\large\frac{11}{2}$

Now $|\overrightarrow {AB}|=3\Rightarrow\:(x+4)^2+y^2=9$

Substituting the value of $x$ we get $y^2=9-\large\frac{9}{4}$

$\Rightarrow\:y=\large\frac{3\sqrt 3}{2}$

$\therefore $ The required displacement $\overrightarrow {OB}=-\large\frac{11}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$