Answer:
Option A
Explanation:
Let x% of the link be used for network.
Given , central frequency =25 GHz
= $25 \times 10^{6} Hz$
Bandwidth of channels = x% of 25 GHz
= $\frac{x}{100} \times 25 \times 10^{9}=25 x \times 10^{7}$
Number of channel= Total bandwidth/ Bandwidth needed per channel
$5 \times 10^{5}=\frac{25 x \times 10^{7}}{2 \times 10^{3}}$
$10 \times 10^{8}=25x \times 10^{7}$
25x=100
$\Rightarrow$ x=4%