Given, a plane electromagnetic wave propagating in the x-direction.
Wavelength of EM wave = 10 nm
Direction of electric field = y direction.
Magnitude of electric field = 60 V/m

Thus, equations for electric and magnetic field is given by,        
                  $$\begin{gathered} {\mathrm{E}}_{\mathrm{y}} = {\mathrm{E}}_{0\mathrm{y}} \sin \left[\frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{ct} - \mathrm{x})\right] \\ {\mathrm{B}}_{\mathrm{z}} = {\mathrm{B}}_{0\mathrm{z}} \sin \left[\frac{2\mathrm{\pi }}{\mathrm{\lambda }}(\mathrm{ct}-\mathrm{x})\right] \end{gathered}$$ 
Using this, a relation between their magnitude is derived as, 
                    ${\mathrm{B}}_{0\mathrm{z}} = \frac{{\mathrm{E}}_{0\mathrm{y}}}{\mathrm{c}}$

Now, on substituting values , we get 

                  $$\begin{gathered} {\mathrm{B}}_{0\mathrm{z}} = \frac{60 \mathrm{v}/\mathrm{m}}{3 \times {10}^8 \mathrm{m}/\mathrm{s}} \\ = 2 \times {10}^{-7}\mathrm{T}. \end{gathered}$$ 

which is the required maximum value of magnetic field.

Given,
Speed of electromagnetic waves, v = 2 $\times$ 10⁸ m/s
Relative permeability of the medium = 1
Speed of light, $\mathrm{C} = 3 \times {10}^8 \mathrm{m}/\mathrm{s}$ 

Using the formula of speed of electromagnetic wave in a medium,
                        $\mathrm{v} = \frac{1}{\sqrt{\mathrm{\mu \epsilon }}}$
                        $\mathrm{v} = \frac{1}{\sqrt{{\mathrm{\mu }}_0{\mathrm{\mu }}_{\mathrm{r}}\left({\mathrm{\epsilon }}_0{\mathrm{\epsilon }}_{\mathrm{r}}\right)}}$
                        $\mathrm{v} = \frac{1}{\sqrt{{\mathrm{\mu }}_0 {\mathrm{\epsilon }}_0}} \times \frac{1}{\sqrt{{\mathrm{\mu }}_{\mathrm{r}} {\mathrm{\epsilon }}_{\mathrm{r}}}}$

Therefore,
Relative permittivity,
                         $$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{r}} = \frac{{\mathrm{c}}^2}{{\mathrm{v}}^2 {\mathrm{\mu }}_{\mathrm{r}}} \\ = \frac{{\left(3 \times {10}^8\right)}^2}{{\left(2 \times {10}^8\right)}^2 \times 1} \\ = 2.25 \end{gathered}$$