Solution:

$z_0=R_0=\sqrt{\frac{L}{C}}=50 \Omega \\ $

with constant $-k$ section low pass filter,

$ \begin{aligned} \\ & W_C=\frac{2}{\sqrt{L C}}=2 \pi \mathrm{fC}_C \\\\ \therefore & 2 \pi \times 50 M H_2=\frac{2}{\sqrt{L C}} \Rightarrow 50 \pi \times 10^6=\frac{1}{\sqrt{L C}}\\ \\ \therefore \quad & \sqrt{L C}=\frac{1}{50 \pi \times 10^6} \quad \therefore L C=4.05 …