Initial velocity $=u$.

Velocity after covering a distance of $x \mathrm{m}=\frac{2\mathrm{u}}{3}$.

For first part of the motion, we can write

$$\begin{gathered} v^2=u^2+2as \\ \Rightarrow {\left(\frac{2u}{3}\right)}^2=u^2-2ax \\ \Rightarrow 2ax=\frac{5u^2}{9} ...Eq\left(1\right) \end{gathered}$$

For the second part of the motion when bullet comes to rest, we can write

$$\begin{gathered} 0^2={\left(\frac{2u}{3}\right)}^2-2a{x}^{\text{'}} \\ \Rightarrow 2a{x}^{\text{'}}=\frac{4u^2}{9} ...Eq\left(2\right) \end{gathered}$$

Dividing equation(2) by equation(1), we get

$\frac{x^{\text{'}}}{x}=\frac{4}{5}$.