First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.

The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.

Setting initial momentum equal to final momentum:

$m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}$

Solving for ($v_{\text{final}}$):

$v_{\text{final}} = \frac{m_{\text{bullet}} \cdot v_{\text{bullet}}}{m_{\text{bullet}} + m_{\text{block}}}$

Substituting the given values:

$v_{\text{final}} = \frac{0.1 \, \text{kg} \cdot 400 \, \text{m/s}}{0.1 \, \text{kg} + 3.9 \, \text{kg}} = 10 \, \text{m/s}$

Next, we know the block comes to rest after moving 20 m due to friction. The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0). The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).

So, setting the work done by friction equal to the initial kinetic energy of the block:

$\mu \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d = \frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2$

Solving for ($\mu$):

$\mu = \frac{\frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2}{(m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d}$

Substituting the given values:

$\mu = \frac{\frac{1}{2} \cdot (0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot (10 \, \text{m/s})^2}{(0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m}} = 0.25$