A bullet of mass 20 g travelling horizontally at a speed of 500 m/s...

Let the mass of the bullet be $m = 20\text{ g} = 0.02\text{ kg}$ and its initial speed be $u = 500\text{ m/s}$.
Let the mass of the block be $M = 10.0\text{ kg}$, which is initially at rest ($U = 0$).
After passing through the block, the bullet's final speed is $v = 100\text{ m/s}$.
Let the velocity of the block immediately after the collision be $V$.

Applying the principle of conservation of linear momentum:
$m u + M U = m v + M V$
$0.02 \times 500 + 10 \times 0 = 0.02 \times 100 + 10 \times V$
$10 = 2 + 10 V$
$10 V = 8 \Rightarrow V = 0.8\text{ m/s}$

The block then slides a distance $s = 20\text{ cm} = 0.2\text{ m}$ before coming to rest. The retardation is provided by the frictional force.
Using the work-energy theorem or kinematic equations:
$V^2 - U^2 = 2as$
$0^2 - (0.8)^2 = 2(-\mu g)s$
$0.64 = 2 \times \mu \times 10 \times 0.2$
$0.64 = 4 \mu$
$\mu = \dfrac{0.64}{4} = 0.16$

The friction coefficient between the block and the surface is $0.16$.