An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a dielectric medium of relative permittivity e_r = 4.0. Then 

When a wave travels from one medium into another, its frequency never changes

For vacuum: $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=v{\lambda }_0 \left(1\right)$

For dielectric medium  $\mathrm{v}=\frac{1}{\sqrt{\mathrm{\mu \epsilon }}}=\mathrm{v\lambda } \left(2\right)$

Dividing (2) by (1) we get

$\frac{\mathrm{\lambda }}{{\mathrm{\lambda }}_0}=\sqrt{\frac{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}{\mathrm{\mu \epsilon }}}$

For non-magnetic media  $\mu ={\mu }_0$. Hence

$$\begin{gathered} \frac{\mathrm{\lambda }}{{\mathrm{\lambda }}_0}=\sqrt{\frac{{\mathrm{\epsilon }}_0}{\mathrm{\epsilon }}}=\sqrt{\frac{1}{{\mathrm{\epsilon }}_{\mathrm{r}}}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\ \therefore \mathrm{\lambda }=\frac{{\mathrm{\lambda }}_0}{2} \end{gathered}$$

Thus the correct choice is (c).