parallel plate

I suppose the battery connection has been withdrawn after charging the plates. Otherwise the potential of the upper plate would still be at +6 v.

First, assume that the upper plate has been charged positively at +6 v by the battery and the lower plate negatively at potential of 0 v. After the insertion of a metal plate between the two charged plates, the upper side of th einserted plate will be -vely charged and lower side +vely charged. Further assuming the the metal plate is of negligible thicknes, its insertion thus doesn't alter the electric field intensity originally set up by the two plates. Because the upper and lower plates are isolated, charges on them are conserved. The potential of the upper plate would remain the same as before.

You could also prove this result mathematically. The insertion of the metal plate in between the two original plates just resulted in forming two capacitors in series. If the capacitance of the original two-plate capacitor is C, then it becomes now two capacitors, each of capacitance 2C connected in series (half the separation between plates doubles the capacitance). The equivalent capacitance of two capacitors, each with capacitance 2C, in series is equal to C. That is, the insertion of the metal plate doesn't change the capacitance of the set up. From the equation: potential = charge/capacitance, the potential is unchanged given that the charge and capacitance of the two-plate set up are the same as before.