Resolve the first displacement into components.

The first displacement is $20.0 \text{ m}$ east.

$$d_{1x} = 20.0 \text{ m}$$ $$d_{1y} = 0 \text{ m}$$

The second displacement is $20.0 \text{ m}$ at $30^\circ$ north of east.

$$d_{2x} = 20.0 \cos(30^\circ) \text{ m}$$ $$d_{2y} = 20.0 \sin(30^\circ) \text{ m}$$

Using $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ and $\sin(30^\circ) = \frac{1}{2}$:

$$d_{2x} = 20.0 \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \text{ m} \approx 17.32 \text{ m}$$ $$d_{2y} = 20.0 \left(\frac{1}{2}\right) = 10.0 \text{ m}$$

$$R_x = d_{1x} + d_{2x} = 20.0 \text{ m} + 10\sqrt{3} \text{ m} = (20.0 + 10\sqrt{3}) \text{ m}$$ $$R_x \approx 20.0 + 17.32 = 37.32 \text{ m}$$ $$R_y = d_{1y} + d_{2y} = 0 \text{ m} + 10.0 \text{ m} = 10.0 \text{ m}$$

The magnitude $R$ is given by the Pythagorean theorem:

$$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(20.0 + 10\sqrt{3})^2 + (10.0)^2}$$ $$R = \sqrt{(400 + 400\sqrt{3} + 300) + 100}$$ $$R = \sqrt{800 + 400\sqrt{3}}$$

Using $\sqrt{3} \approx 1.732$:

$$R = \sqrt{800 + 400(1.732)}$$ $$R = \sqrt{800 + 692.8}$$ $$R = \sqrt{1492.8}$$ $$R \approx 38.637 \text{ m}$$

Rounding to one decimal place, the magnitude is $38.6 \text{ m}$.

Comparing with the given options:
A. 28.3 m
B. 40.0 m
C. 38.6 m
D. 10.4 m

The closest option is C.

The final answer is $\boxed{\text{C. } 38.6 \text{ m}}$.