As we know that,
The standard equation of magnetic field
$$B = {B_0}\sin \left( {kx - \omega t} \right)$$
And, the given equation is
$$B = 12 \times {10^{ - 8}}\sin \left( {1.20 \times {{10}^7}z - 3.60 \times {{10}^{15}}t} \right)T.$$
On comparing this equation with standard equation (i), we have.
$${B_0} = 12 \times {10^{ - 8}}$$
So, the average intensity of the beam
$$\eqalign{ & {I_{{\text{av}}}} = \frac{1}{2}\frac{{B_0^2}}{{{\mu _0}}} \cdot c = \frac{1}{2} \times \frac{{{{\left( {12 \times {{10}^{ - 8}}} \right)}^2} \times 3 \times {{10}^8}}}{{4\pi \times {{10}^{ - 7}}}} \cr & = 1.72\,W/{m^2} \cr} $$

Speed of light of vacuum $$c = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$   and in another medium $$v = \frac{1}{{\sqrt {\mu \varepsilon } }}$$
$$\eqalign{ & \therefore \frac{c}{v} = \sqrt {\frac{{\mu \varepsilon }}{{{\mu _0}{\varepsilon _0}}}} = \sqrt {{\mu _r}K} \cr & \Rightarrow v = \frac{c}{{\sqrt {{\mu _r}K} }}. \cr} $$