Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if:

Writing Kirchoff's laws for the top and bottom loops,
V1 - i1R1 - (i1 - i2)R2 = 0
V2 - i2R3 + (i1 - i2)R2 = 0
If the current in R2 is zero, then i1 - i2 = 0, which means i1 = i2. Substituting this into the equations above:
V1 - i1R1 = 0
V2 - i1R3 = 0
From these equations, we get:
V1 = i1R1
V2 = i1R3
Dividing these equations gives:
V1/V2 = R1/R3
If V1 = V2, then R1 = R3. However, this alone doesn't guarantee i2 = 0.
Let's consider the case where i2 = 0. This means there is no current flowing through R3. Applying Kirchoff's voltage law to the outer loop containing V1, V2, R1, and R3:
V1 - i1R1 - V2 = 0
V1 - V2 = i1R1
If i2 = 0, then i1 flows entirely through R1. The current through R2 would be zero if the potential difference across R2 is zero. This occurs if the potential at the junction between R1 and R2 is the same in both branches. Therefore the voltage across R1 must equal the voltage across R3 for the current in R2 to be zero.
This gives us V1 = V2 and R1 = R3.
Also we know i1R1 = i2R3
Therefore the condition for i2=0 is V1=V2 and R1 = R3.