If the final velocity of the bullet and the bag is $v$, applying conservation of momentum,

$$\begin{gathered} mu=\left(m+M\right)v \\ \Rightarrow 0.2\times 10=\left(0.2+9.8\right)v \\ \Rightarrow v=0.2 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

Loss in kinetic energy will be,

$K_i-K_f=\left(\frac{1}{2}\times 0.2\times {10}^2\right)-\left(\frac{1}{2}\times 10\times 0.2^2\right)=9.8 \mathrm{J}$