Let x be the required length of the slab

The capacitance of a parallel plate capacitor is given by

$C=\frac{{\epsilon }_{0}A}{d}$

Capacitance of parallel plate capacitor before inserting the dielectric slab,
$C_{\text{initial }}=\frac{{\epsilon }_{o}A}{d}\Rightarrow C_{\text{initial }}=\frac{{\epsilon }_{o}lw}{d}$
Now, capacitance of parallel plate capacitor after inserting the dielectric slab
$\begin{array}{l}{C}_{\text{final }}=C_1+C_2\\ =\frac{k{\epsilon }_0{A}_1}{d}+\frac{k{\epsilon }_o{A}_2}{d}=\frac{k{\epsilon }_{o}wx}{d}+\frac{{\epsilon }_{o}w(l-x)}{d}\end{array}$

$=\frac{{\epsilon }_{o}w}{d}[kx+(l-x\left)\right]=\frac{{\omega }_{0}w}{d}[l+(k-1\left)x\right]$

According to question,
Energy stored in capacitor = 2 (initial energy stored)
$\begin{array}{l}\Rightarrow \frac{1}{2}{C}_{\text{final }}{V}^2=2\left(\frac{1}{2}{C}_{\text{initial }}{V}^2\right)\\ \Rightarrow C_{\text{final }}=2C_{\text{initial }}\\ \Rightarrow \frac{{\epsilon }_{o}w}{d}[l+(4-1\left)x\right]=\frac{2{\epsilon }_{0}wl}{d}\\ \Rightarrow \frac{{\epsilon }_{o}w}{d}[l+3x]=\frac{2{\epsilon }_{o}wi}{d}\\ \Rightarrow l+3x=2l\Rightarrow 3x=l\Rightarrow x=\frac{l}{3}\end{array}$