1. Interpretation of the situation

Since after collision the two blocks and the spring move together with constant compression, there is no relative motion between the blocks at that instant. This is the instant of maximum compression of the spring.

So we use:

• Conservation of linear momentum to find common velocity at that instant.
• Conservation of mechanical energy to relate loss of kinetic energy to spring potential energy.

2. Given data

$m_1 = 10\,\text{kg}, \quad m_2 = 5\,\text{kg}, \quad v_1 = 3\,\text{m/s}, \quad v_2 = 0, \quad k = 3000\,\text{N/m}$

Let the common velocity at maximum compression be $V$.

3. Apply conservation of momentum

Initial momentum:

$p_i = m_1 v_1 + m_2 v_2 = 10\times 3 + 5\times 0 = 30$

At maximum compression, both move together:

$p_f = (m_1 + m_2)V = 15V$

So,

$15V = 30$ $V = 2\,\text{m/s}$

4. Apply conservation of energy

Initial kinetic energy:

$K_i = \frac12 m_1 v_1^2 = \frac12 \times 10 \times 3^2 = 45\,\text{J}$

Kinetic energy at maximum compression:

$K_f = \frac12 (m_1+m_2)V^2 = \frac12 \times 15 \times 2^2 = 30\,\text{J}$

Loss in kinetic energy is stored in the spring:

$\frac12 kx^2 = K_i - K_f = 45 - 30 = 15$

Thus,

$\frac12 \times 3000 \times x^2 = 15$ $1500x^2 = 15$ $x^2 = \frac{15}{1500} = 0.01$ $x = 0.1\,\text{m}$

5. Check options

The compression is

$\boxed{0.1\,\text{m}}$

So the correct option is B.