Answer
a) $ v_{\rm bullet} =\dfrac{(m+M) }{m} \sqrt{ 2\mu_kgd } $
b) $443\;\rm m/s$
Work Step by Step
a) We can assume that the system [bullet+block] during the collision time (which is an infinitesimal time) is isolated.
This means that the momentum of the system is conserved.
$$p_i=p_f\\
p_{i,\rm bullet}+p_{i,\rm block}=p_{f,\rm bullet}+p_{f,\rm block}$$
We know that the momentum of the block before the collision is zero since it was at rest.
$$ p_{i,\rm bullet} =p_{f,\rm bullet}+p_{f,\rm block}$$
$$m v_{i,\rm bullet} =(m+M)v_f $$
Thus, $v_{i,\rm bullet} $ is given by
$$ v_{i,\rm bullet} =\dfrac{(m+M)v_f }{m}\tag 1$$
where $v_f$ is the final speed of both of them since the bullet stays inside the block.
Now we need to find their speed just after the collision and to do so we need to use the second part of their motion which is from the collision moment to the final point after a distance of $d$ when they came to rest.
So, we need to draw the force diagram of the system to find its acceleration.
Applying Newton's second law;
$$\sum F_y=F_n-(m+M)g=ma_y=m(0)=0$$
Thus,
$$ F_n=(m+M)g \tag 2$$
$$\sum F_x=-f_k =(m+M)a_x $$
and we know that the friction force is given by $\mu F_n$, so
$$ -\mu_kF_n=(m+M)a_x $$
Plugging from (2);
$$ -\mu_k(m+M)g =(m+M)a_x $$
Hence,
$$a_x= -\mu_k g \tag 3$$
Now we know that the block-bullet system is moving at a constant acceleration, so its velocity just after the collision is given by
$$v_{fx}^2=v_{ix}^2+2a_x\Delta x$$
where $v_{ix}$ is $v_{f}$ in equation (1), $v_{fx}=0$ since they finally stops, and $\Delta x=d$.
$$0^2=v_{f}^2 +2a_xd$$
$$v_f^2=-2a_xd$$
Plugging from (3);
$$v_f^2=-2(-\mu_kg)d$$
Hence,
$$v_f =\sqrt{ 2\mu_kgd }$$
Plugging into (1);
$$ \boxed{ v_{\rm bullet} =\dfrac{(m+M) }{m} \sqrt{ 2\mu_kgd } }$$
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b) In this part, we just need to plug the given into the boxed formula above;
$$ v_{\rm bullet} =\dfrac{(0.010+10) }{0.01} \sqrt{ 2\times 0.2\times9.8\times 0.05 } $$
$$ v_{\rm bullet} =\color{red}{\bf 443}\;\rm m/s$$
