The velocity of the bullet becomes one third after it penetrates...

According to the given information for the first $4cm$, $u = {v}_{0}$ and $v = \dfrac{{v}_{0}}{3}$. Applying equation of motion for constant acceleration,

${v}^{2} = {u}^{2} + 2as$

$⇒ {\left(\dfrac{{v}_{0}}{3}\right)}^{2} = {v}^{2} + 2a\left(4\right) ⇒ a = - \dfrac{8{{v}_{0}}^{2}}{9\left(8\right)} = - \dfrac{{{v}_{0}}^{2}}{9}$

When the body stops, $v = 0$

$⇒ 0 = {{v}_{0}}^{2} + 2a\left(4 + x\right)$

$⇒ {{v}_{0}}^{2} = 2\left(\dfrac{{{v}_{0}}^{2}}{9}\right)\left(4 + x\right)$

$⇒ 4 . 5 = 4 + x$

$⇒ x = 0 . 5cm$