where, $P_{\text {avg }}=$ average power generated by source
$=4 \% \text { of } 100 \mathrm{~W}=\frac{4}{100} \times 100 \mathrm{~W}=4 \mathrm{~W}$
$A=$ area to which energy is transmitted
$=4 \pi r^2=4 \pi \times(2)^2=4 \pi \times 4=16 \pi \mathrm{m}^2$
On putting the values into Eq. (i), we have

where, $\varepsilon_0=$ permittivity of free space
and $E_{\mathrm{rms}}=$ root mean square $(\mathrm{rms})$ value of electric field.
$c=$ speed of light in vacuum $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
On putting the values into Eq. (iii), we get
$I_{\mathrm{avg}}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8$
On equating both the Eqs. (ii) and (iv), we get
$\begin{gathered}
\frac{1}{4 \pi}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 \Rightarrow \frac{1}{4 \pi \varepsilon_0 \times 3 \times 10^8}=E_{\mathrm{rms}}^2 \\
E_{\mathrm{rms}}^2=\frac{1}{4 \pi \varepsilon_0} \times \frac{1}{3 \times 10^8}=9 \times 10^9 \times \frac{1}{3 \times 10^8} \\
\quad\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\
=3 \times 10=\sqrt{30} \mathrm{~V} / \mathrm{m}
\end{gathered}$