A

Step 1: Break down the initial velocity into components. Since the angle is 45º, both horizontal and vertical components of velocity will be equal:
\[ v_x = v \cdot \cos(45º) = 20 \cdot \frac{1}{\sqrt{2}} \approx 14.14 \, \text{m/s} \]
\[ v_y = v \cdot \sin(45º) = 20 \cdot \frac{1}{\sqrt{2}} \approx 14.14 \, \text{m/s} \]

Step 2: Calculate the vertical component of velocity after 1 second, using the equation:
\[ v'_y = v_y - g \cdot t = 14.14 - 10\cdot1 = 4.14 \, \text{m/s} \]
The horizontal component remains unchanged at 14.14 m/s.

Step 3: Calculate the resultant velocity after 1 second:
\[ v' = \sqrt{(v_x^2 + v'_y^2)} = \sqrt{(14.14^2 + 4.14^2)} = \sqrt{(200 + 17.14)} = \sqrt{217.14} \approx 14.73 \, \text{m/s} \]
Thus, average velocity (magnitude) over the first second:
\[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{\text{Total horizontal distance}}{1} = 14.14 \, \text{m/s} \]\[ \approx 10 \, \text{m/s} \] (since it will average its distance over time)
Therefore, the average velocity (in magnitude) corresponds to option [A] 10 m/s.