Given:
- BW=$20$Mbps
- Distance=$2500$m
- L=$512$m
- V=$2\times10^8$m/sec
$T_p=\frac{d}{v}\implies\frac{2500m*sec}{2\times10^8m}=12.5\mu sec$
$T_t=\frac{L}{BW}\implies\frac{512 bit*sec}{20\times10^6 bit}=25.6\mu sec$
The efficiency of Ethernet: $\eta=\frac{1}{1+6.44a},\text{where $a=T_p/T_t$}$
Now $a=\frac{T_p}{T_t}=0.488$
$\therefore \eta=\frac{1}{1+6.44*0.488}=24.12\%$
So The percentage of time the channel is idle=$100-24.12=75.88\%$
Ref: Efficiency of Ethernet