For the first part of the motion of the bullet into the block, it can be written that

$\begin{array}{rcl}{\left(\frac{u}{3}\right)}^2& =& u^2-2a\times 24\\ & \Rightarrow & 48a=\frac{8u^2}{9} ...\left(1\right)\end{array}$

For the second part of the motion of the bullet through the block, it can be written that

$\begin{array}{rcl}0& =& {\left(\frac{u}{3}\right)}^2-2aS\\ & \Rightarrow & 2aS=\frac{u^2}{9} ...\left(2\right)\end{array}$

Divide equation (2) by equation (1) to obtain the distance travelled by the bullet during the second part of its motion.

$\begin{array}{rcl}\frac{2aS}{48a}& =& \frac{{\displaystyle \frac{{\displaystyle u^2}}{{\displaystyle 9}}}}{\frac{8u^2}{9}}\\ & \Rightarrow & S=3\end{array}$

Hence, the total length of the block is given by

$\begin{array}{rcl}L& =& \left(3+24\right) \mathrm{cm}\\ & =& 27 \mathrm{cm}\end{array}$