Consider the arrangement depicted in the figure. The capacitor...

Let the width of the capacitor plates be $b$. The area of each plate is $A = bl$, which gives $b = \dfrac{A}{l}$.

When a length $x$ of the dielectric slab is inside the capacitor, the system can be considered as two capacitors in parallel: one with air and one with the dielectric.
The equivalent capacitance is:
$C = \dfrac{\varepsilon_0 b (l - x)}{d} + \dfrac{K \varepsilon_0 b x}{d} = \dfrac{\varepsilon_0 b}{d} [l + (K - 1)x]$

The force exerted by the electric field on the dielectric slab is constant and is given by:
$F = \dfrac{1}{2} \mathcal{E}^2 \dfrac{dC}{dx} = \dfrac{1}{2} \mathcal{E}^2 \left[ \dfrac{\varepsilon_0 b (K - 1)}{d} \right] = \dfrac{\varepsilon_0 A (K - 1) \mathcal{E}^2}{2dl}$

Since the force is constant, the acceleration of the slab is also constant:
$a_c = \dfrac{F}{m} = \dfrac{\varepsilon_0 A (K - 1) \mathcal{E}^2}{2mdl}$

The slab is released from rest with an initial length $a$ inside the capacitor. It will accelerate until it is completely inside the capacitor. The distance it moves during this acceleration phase is $(l - a)$.
Using the equation of motion $s = \dfrac{1}{2} a_c t_1^2$, the time $t_1$ taken to reach the fully inside position is:
$l - a = \dfrac{1}{2} a_c t_1^2$
$\Rightarrow t_1 = \sqrt{\dfrac{2(l - a)}{a_c}} = \sqrt{\dfrac{2(l - a) \times 2mdl}{\varepsilon_0 A (K - 1) \mathcal{E}^2}} = 2\sqrt{\dfrac{mdl(l - a)}{\varepsilon_0 A (K - 1) \mathcal{E}^2}}$

Due to inertia, the slab will overshoot and start emerging from the other side. The electric force will now act in the opposite direction, decelerating the slab until it momentarily comes to rest after moving a distance $(l - a)$ outside. It will then be pulled back in.
The complete periodic motion consists of four such identical segments (accelerating inwards, decelerating outwards, accelerating inwards again, decelerating outwards to the initial position).

Therefore, the total time period $T$ of the oscillation is:
$T = 4t_1 = 8\sqrt{\dfrac{mdl(l - a)}{\varepsilon_0 A \mathcal{E}^2 (K - 1)}}$

Answer: $8\sqrt{\dfrac{(l - a) l m d}{\varepsilon_0 A \mathcal{E}^2 (K - 1)}}$