To solve this problem, we start by using the conservation of linear momentum.

Initially, block A of mass $ m_1 = 10 \, \text{kg} $ is moving with velocity $ v_1 = 3 \, \text{m/s} $, while block B of mass $ m_2 = 5 \, \text{kg} $ is at rest. The blocks move together after the collision. The final velocity of the system is $ v_{\text{cm}} $. Applying the conservation of linear momentum, we have:

$ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{cm}} $

Substituting the known values:

$ 10 \times 3 + 5 \times 0 = (10 + 5) v_{\text{cm}} $

$ 30 = 15 v_{\text{cm}} $

Solving for $ v_{\text{cm}} $:

$ v_{\text{cm}} = \frac{30}{15} = 2 \, \text{m/s} $

Next, we use the conservation of energy to find the compression in the spring. The initial kinetic energy of block A is given by:

$ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 10 \times 3^2 = 45 \, \text{J} $

The final kinetic energy of both blocks moving together at $ v_{\text{cm}} = 2 \, \text{m/s} $ is:

$ \frac{1}{2} (m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2} \times 15 \times 2^2 = 30 \, \text{J} $

The difference in kinetic energy is the energy stored in the compressed spring:

$ \frac{1}{2} k x^2 = 45 - 30 = 15 \, \text{J} $

Substituting the given spring constant $ k = 3000 \, \text{N/m} $ into the energy equation:

$ \frac{1}{2} \times 3000 \times x^2 = 15 $

$ 1500 x^2 = 15 $

Solving for $ x^2 $:

$ x^2 = \frac{15}{1500} = \frac{1}{100} $

Finally, solving for $ x $:

$ x = \frac{1}{10} = 0.1 \, \text{m} $

Therefore, the compression in the spring is $ 0.1 \, \text{m} $.