Answer
$$7.5 \mathrm{\ cm}$$
Work Step by Step
Using unit-vector notation, the displacement vector for point $A$ is
$\vec{d}_{A}=\vec{w}+\vec{v}+\vec{i}+\vec{h} \\ \quad \ \ =l_{0}\left(\cos 60^{\circ} \hat{\mathrm{i}}+\sin 60^{\circ} \hat{\mathrm{j}}\right)+\left(l_{0} \hat{\mathrm{j}}\right)+l_{0}\left(\cos 120^{\circ} \hat{\mathrm{i}}+\sin 120^{\circ} \hat{\mathrm{j}}\right)+\left(l_{0} \hat{\mathrm{j}}\right)$
Therefore, the magnitude of $\vec{d}_{A}$ is $\left|\vec{d}_{A}\right|=(2+\sqrt{3})(2.0 \mathrm{cm})=7.5 \mathrm{cm}$
