Introduction:
In this problem, we need to calculate the capacitance of a parallel plate condenser when a glass plate of dielectric constant 9 is introduced between the plates. The distance between the plates is given as 6mm and the thickness of the glass plate is given as 4.5mm.

Solution:
The capacitance of a parallel plate condenser is given by the formula:

C = εA/d

where ε is the permittivity of the medium between the plates, A is the area of the plates, and d is the distance between the plates.

Step 1: Calculate the initial capacitance of the parallel plate condenser.

Initial distance between plates, d = 6mm = 0.006m
Area of the plates, A = ?
As the area of the plates is not given, we assume it to be 1m2 for simplicity.

The permittivity of free space, ε0 = 8.85 x 10^-12 F/m

The permittivity of the medium between the plates, εr = 1 (as no medium is mentioned)

Using the formula,
C = εA/d = ε0 x A/d = 8.85 x 10^-12 x 1 / 0.006 = 1.475 x 10^-9 F

Hence, the initial capacitance of the parallel plate condenser is 1.475 x 10^-9 F.

Step 2: Calculate the capacitance of the parallel plate condenser when the glass plate is introduced.

Distance between plates, d' = d - t = 0.006 - 0.0045 = 0.0015m (where t is the thickness of the glass plate)
Area of the plates, A = 1m2
The permittivity of free space, ε0 = 8.85 x 10^-12 F/m
The permittivity of the medium between the plates, εr = 9 (as glass has a dielectric constant of 9)

Using the formula,
C' = εA/d' = ε0 x εr x A/d' = 8.85 x 10^-12 x 9 x 1 / 0.0015 = 5.29 x 10^-9 F

Hence, the capacitance of the parallel plate condenser when the glass plate is introduced is 5.29 x 10^-9 F.

Conclusion:
Therefore, the capacitance of the parallel plate condenser increases from 1.475 x 10^-9 F to 5.29 x 10^-9 F when a glass plate of dielectric constant 9 and thickness 4.5mm is introduced between the plates.