In Ethernet the Transmission Time should be greater than equal to twice that of the propagation time.(Ethernet uses CSMA/CD).
For network 1: L(length of the data)=x bits For network 2:L(length of the data)=y bits
distance i.e d=3 km distance i.e d=6 km
v=2/3 * velocity of light v=1/3*velocity of light
so putting these values in the above equation we obtain x=300 bits and y=24000bits
So the answer i.e. x+y=24000+300=24300 bits.