Time when second child leaves (T2) = 1:15 pm
Speed of second child (S2) = 6 mi/hr

Time elapsed until second child leaves=T2-T1=15 minutes
Distance travelled by first child in that time=s1*15 minutes
=4 mi/hr * (1/4)hr
=1 mile

Range of radio=2 miles
Distance between children at 1.15 pm = 1 mile
Remaining range=1 mile

first child and second child are moving in opposite directions hence both distances must be considered

s=d/t => d=t*s

(distanceNorth)+(distanceSouth)=1 mile
t(S1)+t(S2)=1 mile
t(S1+S2)=1 mile
t(4+6)=1
t(10)=1
t=1/10 hour=60/10 minutes=6 minutes

Thus,10 minutes later the remaining range of 1 mile would also be passed.
Time at that moment = 1.15+6 minutes = 1:21 pm

Ans: After 1:21 pm,they will be unable to communicate with each other.