If $u$is the initial velocity of bullet then after passing through one plank the velocity became $v=u-\frac{u}{25}=\frac{24u}{25}$

Using kinematic equation for uniform acceleration motion, $v^2-u^2=2as$

$$\begin{gathered} {\left(\frac{24u}{25}\right)}^2-u^2=2as \\ \Rightarrow 2as=\frac{u^2\left(-625+576\right)}{625} \\ \Rightarrow 2as=-u^2\frac{49}{625} \end{gathered}$$

If $n$number of planks are required to bring the bullet at rest then, $v=0$

$$\begin{gathered} \Rightarrow 0-u^2=2a\left(ns\right) \\ \Rightarrow -u^2=n\left(\frac{-49u^2}{625}\right) \\ \Rightarrow n=\frac{625}{49}=12.75\text{ or }13 \end{gathered}$$

Thus, $13$planks will be required in order to bring the bullet at rest.