C

Step 1: Resolve each vector into its x and y components.
For vector A (magnitude 5N at 0°):
A_x = 5 cos(0°) = 5N, A_y = 5 sin(0°) = 0N.
For vector B (magnitude 3N at 90°):
B_x = 3 cos(90°) = 0N, B_y = 3 sin(90°) = 3N.
For vector C (magnitude 4N at 180°):
C_x = 4 cos(180°) = -4N, C_y = 4 sin(180°) = 0N.

Step 2: Calculate the resultant vector by adding the components.
R_x = A_x + B_x + C_x = 5 + 0 - 4 = 1N.
R_y = A_y + B_y + C_y = 0 + 3 + 0 = 3N.

Step 3: Find the magnitude of the resultant vector using Pythagoras' theorem:
R = \sqrt{R_x^2 + R_y^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16 N.

Step 4: Find the direction using trigonometry:
\theta = \tan^{-1}(R_y/R_x) = \tan^{-1}(3/1) \approx 71.57°.

Therefore, the resultant vector has a magnitude of approximately 3.16 N at an angle of about 71.57° from the positive x-axis, corresponding to option C.