BJT Circuit Base-Emmitter voltage
Here is my explanation (calculation):
V3=Vout+0.7V with Vout=V1=0.7V

V3=(0.7+0.7)V=1.4V .

This result must be in accordance with Ohms law for R2:
V3=3-I2R2=3-0.5mA*3.2kohms=3-1.6=1.4V.

This calculation shows that both results for V3 are equal for Ic=0.5mA only.
Because this is half of the emitter current source both transistors draw the same current.
Hence, both base-emitter voltages must be equal.

EDIT/UPDATE:

To answer the question finally (the task was to find the relationshipo between V2 and VB) - here is a corresponding formula:

V2=VB + VT*ln{[IE2*R2/(Vcc-2*VBE)-1]}

Rough calculation (with VT=26mV, IE2=1mA, R2=3.2k, VBE=0.7V):

V2=VB + VT*ln[(3.2/1.6)-1)

V2=VB .