Don't try this at home!

$child's\quad velocity=-6m/s\\ (Velocity\quad towards\quad right=+ve,\quad towards\quad left=-ve)\\ \\ 1.\quad Jumping\quad From\quad B\\ Momentum\quad Before\quad jump=Momentum\quad After\quad Jump\\ 0=4(-6)+20({ v }'_{ b })\\ { v }'_{ b }=\frac { 6 }{ 5 } m/s\\ 2.\quad Landing\quad on\quad A\quad then\quad Jumping\quad from\quad A.\\ Momentum\quad Just\quad Before\quad landing\quad on\quad A=\\ Momentum\quad Just\quad After\quad Jumping\quad from\quad A\\ 4(-6)=4(v)+20({ v }_{ a })\\ but\quad given\quad v-{ v }_{ a }=6\quad (Jumps\quad 6m/s\quad relative\quad to\quad the\quad cart)\\ \qquad \qquad \quad \quad v=6+{ v }_{ a }\\ Substituting\quad we\quad get\quad v=4m/s,\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { v }_{ a }=-2m/s\\ 3.\quad Landing\quad on\quad B\\ Momentum\quad Before\quad landing=Momentum\quad After\quad landing\\ 4(4)+20(\frac { 6 }{ 5 } )=24({ v }_{ b })\quad (They\quad move\quad together\quad therefore\quad v={ v }_{ b })\\ { v }_{ b }=\frac { 5 }{ 3 } m/s\\ \frac { 6{ v }_{ b } }{ 5{ v }_{ a } } =\frac { 6(\frac { 5 }{ 3 } ) }{ 5(2) } \quad =1$

Consider the system to be "cart A+cartB+child". Initial momentum p=0. After the jumps, mass of cart B=24kg, mass of cart A=20 kg. Thus, final momentum P=24V B - 20V A. By law of momentum conservation, equating p=P, we have the required ratio=1