A parallel plate capacitor with plate separation 5mm is charged up...

Without dielectric, capacitance will be $C = \dfrac{A{ε}_{0}}{d}$.

Therefore, charge on it, $Q = \dfrac{A{ε}_{0}}{d}V$

Now, with dielectric capacitance will be ${C}^{'} = \dfrac{A{ε}_{0}}{d - t + \dfrac{t}{K}}$.

Hence, charge on it will be ${Q}^{'} = \dfrac{A{ε}_{0}V}{d - t + \dfrac{t}{K}}$

Given:

$\dfrac{A{ε}_{0}V}{d - t + \dfrac{t}{K}} = \left(1 . 25\right)\dfrac{A{ε}_{0}V}{d}$

$⇒ 1 . 25\left(3 + \dfrac{2}{K}\right) = 5$

$⇒ K = 2$