A thin paper of thickness 0.02 mm with a refractive index of 1.45...

Let the intensity of light from the unblocked slit be $I_0$.

Since the paper transmits $4/9$ of the incident light energy, the intensity of light from the blocked slit is $I_1 = \dfrac{4}{9}I_0$.

The maximum intensity in the interference pattern is given by:

I_\max = \left(\sqrt{I_0} + \sqrt{I_1}\right)^2

I_\max = \left(\sqrt{I_0} + \sqrt{\dfrac{4}{9}I_0}\right)^2 = \left(1 + \dfrac{2}{3}\right)^2 I_0 = \dfrac{25}{9}I_0

The minimum intensity in the interference pattern is given by:

I_\min = \left(\sqrt{I_0} - \sqrt{I_1}\right)^2

I_\min = \left(\sqrt{I_0} - \sqrt{\dfrac{4}{9}I_0}\right)^2 = \left(1 - \dfrac{2}{3}\right)^2 I_0 = \dfrac{1}{9}I_0

The ratio of the maximum intensity to the minimum intensity is:

\dfrac{I_\max}{I_\min} = \dfrac{\dfrac{25}{9}I_0}{\dfrac{1}{9}I_0} = 25

The thickness and refractive index of the paper introduce a constant phase difference which shifts the fringe pattern but does not affect the values of maximum and minimum intensities.