Energy density of an EMwave =21εE02, whare E0 is the amplitude of the wave. Since total energy is same for both cylinders $\begin{aligned}
& \left(\frac{1}{2} \varepsilon E_1^2\right) \pi R_1^2 L_1=\left(\frac{1}{2} \varepsilon E_2^2\right) \pi R_2^2 L_2 \
& \Rightarrow E_1^2 R_1^2 L_1=E_2^2 R_2^2 L_2 \
& \text { or } E_2=\frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}}=\frac{100 \mathrm{d}}{(\mathrm{d} / 2)} \sqrt{\frac{L_1}{L 1}}=200 \mathrm{N} / \mathrm{C} \
& \qquad \quad\left[\because L_1=L_2=200 \mathrm{cm}\right]
\end{aligned}\RightarrowTheamplitudeofcorrespondingE Mwaveis200N/CorthewaveisE=200 \sin (\omega t-k x) \mathrm{NC}^{-1}$