A

Step 1: Determine the spring constant, k. The force due to the weight of the block is given by \( F = mg \), where \( m = 10 \text{ kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Therefore, \( F = 10 \times 9.8 = 98 \, \text{N} \).
Step 2: At equilibrium, the force exerted by the spring is equal to the weight of the block. By Hooke's Law, \( F = k \Delta x \), where \( \Delta x = 0.2 \, \text{m} \). Thus, \( k \times 0.2 = 98 \) which gives \( k = \frac{98}{0.2} = 490 \, \text{N/m} \).
Step 3: When the block is given an upward speed of 4 m/s, we calculate the maximum height it reaches above the equilibrium position using the conservation of energy. The initial kinetic energy is given by \( KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 10 \times (4^2) = 80 \, \text{J} \).
Step 4: At the maximum height, all kinetic energy is converted into potential energy stored in the spring, given by \( PE = \frac{1}{2} k x^2 \), where \( x \) is the stretch of the spring from the equilibrium position.
Equating the energies, we have:
\( 80 = \frac{1}{2} \times 490 \times x^2 \).
Step 5: Rearranging gives \( x^2 = \frac{160}{490} \), and taking the square root gives \( x = \sqrt{\frac{160}{490}} \approx 0.569 \, \text{m} \) or 56.9 cm.
Therefore, the maximum height the block will rise from the equilibrium position is approximately 0.569 m.
Thus, the answer is A.