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

The path difference introduced by a transparent sheet of thickness $t$ and refractive index $\mu$ is given by $\Delta x = (\mu - 1)t$.

The number of fringes shifted due to this path difference is $n = \dfrac{\Delta x}{\lambda} = \dfrac{(\mu - 1)t}{\lambda}$.

Substituting the given values:

$n = \dfrac{(1.45 - 1) \times 0.02 \times 10^{-3}}{600 \times 10^{-9}}$

$n = \dfrac{0.45 \times 2 \times 10^{-5}}{6 \times 10^{-7}} = \dfrac{9 \times 10^{-6}}{6 \times 10^{-7}} = 15$

When the first paper is pasted, the central fringe shifts by $15$ fringes. When an identical paper is subsequently pasted across the second slit, the new path difference exactly cancels the initial one, bringing the central maximum back to the geometric centre.

Thus, the number of fringes that cross the centre during this second process is $15$. The transmission of $4/9$ of the incident light energy only affects the intensity and contrast of the fringes, not their positions.