An antenna is placed in a dielectric medium of dielectric constant...

We know that, refractive index is related to velocity as $n = \dfrac{c}{v}$.

Here, $n = \sqrt{{ε}_{r}{\mu}_{r}} = \sqrt{\left(6 . 25\right)\left(1\right)} = 2 . 5$

Since, effective length of antenna $l > \dfrac{\lambda}{4} ⇒ l > \dfrac{v}{4f}$

So, $f > \dfrac{v}{4l} = \dfrac{c}{4l\left(2 . 5\right)} = \dfrac{3 \times {10}^{8}}{2 . 5 \times 4 \times 5 \times {10}^{- 3}} = 6 \times {10}^{9}Hz = 6GHz$.