Given,

Length of plate of parallel plate capacitor is $l$

Width of plate is $w$

Separation of plates is $d$.

Emf of battery is $V$

Thickness of dielectric slab is $d$

Dielectric constant of slab is $K = 4$

Capacitance of capacitor before the insertion of dielectric slab is 

$C_0=\frac{A{\epsilon }_0}{d}=\frac{wl{\epsilon }_0}{d} ...\left(1\right)$

Energy stored in the capacitor before the insertion of dielectrics is given by,

$U_0=\frac{1}{2}{C}_0{V}^2=\frac{1}{2}\left(\frac{wl{\epsilon }_0}{d}\right)V^2 ...\left(2\right)$

Suppose that, the slab is inside the plate at $y$ distance, therefore, it becomes the combination of two capacitors one with air with length $(l-y)$ and another with dielectric with length $y$ as shown in diagram.

Now the capacitance of capacitor in parallel combination is given by,

$C=C_1+C_2 ...\left(3\right)$

Where,

$C_1=\frac{w(l-y){\epsilon }_0}{d} ...\left(4\right)$

$C_1=\frac{wyK{\epsilon }_0}{d} ...\left(5\right)$

Now energy stored in this combination of capacitor is given by

$U=\frac{1}{2}(C_1+C_2)V^2 ...\left(6\right)$

Now according to the given condition,

$2U_0=U$

$\Rightarrow 2\left(\frac{1}{2}{C}_0{V}^2\right)=\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){\mathrm{V}}^2$

$\Rightarrow 2{\mathrm{C}}_0={\mathrm{C}}_1+{\mathrm{C}}_2$

$\Rightarrow 2\left(\frac{{\mathit{ϵ}}_{\mathit{0}}\mathit{w}\mathit{l}}{\mathit{d}}\right)=\frac{{\mathit{ϵ}}_{\mathit{0}}\mathit{K}\mathit{w}\mathit{y}}{\mathit{d}}+\frac{{ϵ}_{\mathit{0}}w(l-\mathrm{y})}{\mathrm{d}}$

$\Rightarrow 2l=Ky+l-\mathrm{y}$

$\Rightarrow l=4y-y=3y$

$\Rightarrow \mathrm{y}=\frac{l}{3}$