Answer
the magnitude of $\vec{d}_{B}$ is
$$=8.6 \mathrm{\ cm}$$
Work Step by Step
Similarly, the displacement for point $B$ is
$\vec{d}_{B}=\vec{w}+\vec{v}+\vec{j}+\vec{p}+\vec{o}$
$=l_{0}\left(\cos 60^{\circ} \hat{\mathrm{i}}+\sin 60^{\circ} \hat{\mathrm{j}}\right)+\left(l_{0} \hat{\mathrm{j}}\right)+l_{0}\left(\cos 60^{\circ} \hat{\mathrm{j}}+\sin 60^{\circ} \hat{\mathrm{j}}\right)+l_{0}\left(\cos 30^{\circ} \hat{\mathrm{j}}+\sin 30^{\circ} \hat{\mathrm{j}}\right)+\left(l_{0} \hat{\mathrm{i}}\right)$
$=(2+\sqrt{3} / 2) l_{0} \hat{\mathrm{i}}+(3 / 2+\sqrt{3}) l_{0} \hat{\mathrm{j}}$
Therefore, the magnitude of $\vec{d}_{B}$ is
$\left|\vec{d}_{B}\right|=l_{0} \sqrt{(2+\sqrt{3} / 2)^{2}+(3 / 2+\sqrt{3})^{2}}=(2.0 \mathrm{cm})(4.3)=8.6 \mathrm{cm}$
