Step 1:

A girl walks 4km toward west (i.e) along $OX$ axis.

Therefore $\overrightarrow{OP}=-4\hat i$----(1)

Next she goes in the direction $30^{\large\circ}$ east of north (i.e) she moves along $OQ$ and stops at $Q$

This implies $PQ$ makes an angle $60^{\large\circ}$ with $OP$.

Step 2:

Hence the scalar component of $PQ$ along $OX=OQ\cos 60^{\large \circ}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=3\cos60^{\large\circ}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=3\times \large\frac{1}{2}=\frac{3}{2}$

Step 3:

Scalar vertical component of $PQ$ along $OY=OQ\sin 60^{\large\circ}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=3\sin 60^{\large\circ}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{3\sqrt 3}{2}$

Therefore $\overrightarrow{PQ}=\large\frac{3}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$------(2)

Step 4:

The girl walks along $OP$ and then along $PQ$

$\overrightarrow {OP}+\overrightarrow{PQ}=\overrightarrow{OQ}$ (By triangle law of vectors)

By adding eq(1) and eq(2) we get

$\overrightarrow{OQ}=-4\hat i+\large\frac{3}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$

$\overrightarrow{OQ}=-4\hat i+\large\frac{3}{2}$$\hat i+\big(\large\frac{3\sqrt 3}{2}$$\hat j\big)$

$\quad\quad=\large\frac{5}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$

$|\overrightarrow {OQ}|=\sqrt{\big(\large\frac{5}{2}\big)^2+\big(\frac{3\sqrt 3}{2}\big)^2}$

$\qquad=\sqrt{\big(\large\frac{25}{4}\big)+\big(\large\frac{27}{4}\big)}$

$\qquad=\sqrt{\big(\large\frac{52}{4}\big)}$

$\qquad=\sqrt{13}$

Step 5:

Hence the distance travelled by the girl from her initial position is $\sqrt{13}$ units.