(1)

Given, $E_{\text{rms}}=6$$\text{V}$${\mathrm{m}}^{-1}$

$\frac{E_{\text{rms}}}{B_{\text{rms}}}=c \text{or} B_{\text{rms}}=\frac{E_{\text{rms}}}{c}\Rightarrow B_{\text{rms}}=\frac{6}{3\times {10}^8}=2\times {10}^{-8}\text{T}$

Since, $B_{\text{rms}}=\frac{B_0}{\sqrt{2}}$

where $B_0$ is the peak value of magnetic field.

$\therefore$   $B_0=B_{\text{rms}}\sqrt{2}=2\times {10}^{-8}\times \sqrt{2}\text{T}$

        $B_0\approx 2.83\times {10}^{-8}\text{T}$

(2)

An accelerating charge is used to produce oscillating electric and magnetic fields, hence the electromagnetic wave.

(2)

Here, Energy flux, $I=25\times {10}^4$$\text{W}$${\mathrm{m}}^{-2}$

Area, $A=15$${\text{cm}}^2=15\times {10}^{-4}{\text{m}}^2$

Speed of light, $c=3\times {10}^8{\text{ms}}^{-1}$

For a perfectly reflecting surface, the average force exerted on the surface is

$F=\frac{2IA}{c}=\frac{2\times 25\times {10}^4 {\text{Wm}}^{-2}\times 15\times {10}^{-4}{\text{m}}^2}{3\times {10}^8{\text{m s}}^{-1}}$

                 $=250\times {10}^{-8}\text{N}=2.50\times {10}^{-6}\text{N}$