b
\(\begin{array}{l}
Given,\,{m_1} = 4g,{u_1} = 300m/s\\
{m_2} = 0.8\,kg = 800\,g,\,{u_2} = 0\,m/s\\
From\,law\,of\,conservation\,of\,momentum,\\
{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\\
Let\,the\,velocity\,of\,combined\,system\\
 = v\,m/s\\
then,\\
4 \times 300 + 800 \times 0 = \left( {800 + 4} \right) \times v\\
v = \frac{{1200}}{{804}} = 1.49\,m/s
\end{array}\)

\(\begin{array}{l}
Now,\,\mu  = 0.3\left( {given} \right)\\
a = \mu g\\
a = 0.3 \times 10\left( {take\,g = 10\,m/{s^2}} \right)\\
 = 3\,m/{s^2}\\
then,\,from\,{v^2} = {u^2} + 2as\\
{\left( {1.49} \right)^2} = 0 + 2 \times 3 \times s\\
s = \frac{{\left( {{{1.49}^2}} \right)}}{6}\,;\,\,s = \frac{{2.22}}{6} = 0.379\,m
\end{array}\)