Based on the figure, determine the magnitude of the resultant of OA...

Let us resolve each vector into its $x$ and $y$ components.

For vector $\vec{OA}$:
Magnitude $= 2.0 \text{ m}$, angle with positive X-axis = 30^{}^{{}^{\circ}}
\vec{OA} = 2.0 \cos(30^{}^{{}^{\circ}}) \hat{i} + 2.0 \sin(30^{}^{{}^{\circ}}) \hat{j}
$\vec{OA} = 2.0 \left(\dfrac{\sqrt{3}}{2}\right) \hat{i} + 2.0 \left(\dfrac{1}{2}\right) \hat{j} = \sqrt{3} \hat{i} + 1.0 \hat{j}$

For vector $\vec{BC}$:
Magnitude $= 1.5 \text{ m}$, angle with negative X-axis = 60^{}^{{}^{\circ}}
\vec{BC} = -1.5 \cos(60^{}^{{}^{\circ}}) \hat{i} + 1.5 \sin(60^{}^{{}^{\circ}}) \hat{j}
$\vec{BC} = -1.5 \left(\dfrac{1}{2}\right) \hat{i} + 1.5 \left(\dfrac{\sqrt{3}}{2}\right) \hat{j} = -0.75 \hat{i} + 0.75\sqrt{3} \hat{j}$

For vector $\vec{DE}$:
Magnitude $= 1.0 \text{ m}$, directed along negative Y-axis
$\vec{DE} = -1.0 \hat{j}$

The resultant vector $\vec{R}$ is the sum of these three vectors:
$\vec{R} = \vec{OA} + \vec{BC} + \vec{DE}$
$\vec{R} = (\sqrt{3} - 0.75) \hat{i} + (1.0 + 0.75\sqrt{3} - 1.0) \hat{j}$
$\vec{R} = (\sqrt{3} - 0.75) \hat{i} + 0.75\sqrt{3} \hat{j}$

Now, we calculate the magnitude of the resultant vector:
$|\vec{R}| = \sqrt{(\sqrt{3} - 0.75)^2 + (0.75\sqrt{3})^2}$
$|\vec{R}| = \sqrt{3 - 1.5\sqrt{3} + 0.5625 + 0.5625 \times 3}$
$|\vec{R}| = \sqrt{3.5625 - 1.5\sqrt{3} + 1.6875}$
$|\vec{R}| = \sqrt{5.25 - 1.5(1.732)}$
$|\vec{R}| = \sqrt{5.25 - 2.598} = \sqrt{2.652}$
$|\vec{R}| \approx 1.628 \text{ m}$

Rounding to one decimal place, the magnitude is approximately $1.6 \text{ m}$.

Answer: $1.6 \text{ m}$