Initial velocity of bullet $v_{1}=500 \mathrm{m} / \mathrm{s}$

Mass of block $M=10 \mathrm{kg}$

Initial velocity $v_{2}=0$

Final velocity $v_{1}^{\prime}=100 \mathrm{m} / \mathrm{s}$

Now, the final velocity of block when bullet comes out

If block velocity $=v^{\prime}_{2}$

$m v_{1}+M v_{2}=m v_{1}+M v_{2}^{\prime}$

$0.02 \times 500=0.02 \times 100+10 v_{2}^{\prime}$

$v_{2}^{\prime}=\frac{10-2}{10}$

$v_{2}^{\prime}=0.8 \mathrm{m} / \mathrm{s}$

Now, after moving $0.2$ $\mathrm{m}$

Change in kinetic energy $=$ work done

$0-\frac{1}{2} \times 10 \times(0.8)^{2}=-\mu \times 10 \times 10 \times 0.2$

$\mu=0.16$

Hence, the friction coefficient is $0.16$