After penetrating $3\, cm$ its velocity becomes $\frac{u}{2}$

From ${v^2} = {u^2} - 2as$

${\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\,(3)$

$⇒$ $6a = \frac{{3{u^2}}}{4}$ $⇒$  $a = \frac{{{u^2}}}{8}$

Let further it will penetrate through distance $x$ and stops at point $C$.

For distance $BC$, $v = 0,\,u = u/2,\,s = x,\,a = {u^2}/8$

From ${v^2} = {u^2} - 2as$ $⇒$ $0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{{u^2}}}{8}} \right)\,.\,x$ $⇒$ $x = 1\,cm$.