The girl walks $4 \mathrm{~km}$ westward vector
$$
\overline{\mathrm{OP}}=4 \hat{\mathrm{i}}
$$
She goes in the direction $30^{\circ}$ east of north i.e., she moves along OQ and stops at Q.PQ makes an angle of $60^{\circ}$ with OP.


Scalar component of PQ along $\mathrm{OX}=\mathrm{OQ} \cos 60^{\circ}$
$$
=3 \cos 60^{\circ}=\frac{3}{2}
$$
Scalar vertical component of PQ along
$$
\begin{aligned}
&\mathrm{OY}=3 \sin 60^{\circ}=\frac{3 \sqrt{3}}{2} \\
&\therefore \quad \overline{\mathrm{PQ}}=\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}
\end{aligned}
$$
Thus the girl walks along $\mathrm{OP}$ and then along PQ
$$
\begin{aligned}
&\therefore \quad \overline{\mathrm{OP}}+\overline{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}} \\
&\text { from (i) \& (ii) }-4 \hat{\mathrm{i}}+\left(\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\right)=\mathrm{OQ} \\
&\therefore \quad \overline{\mathrm{OQ}}=-\frac{5}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}} \\
&\qquad|\overrightarrow{\mathrm{OQ}}|=\sqrt{13}
\end{aligned}
$$