1. Condition for successive minima in transmitted light

For a thin film of refractive index $\mu=1.4$ in air, at normal incidence:

• One reflected ray gets a phase reversal of $\pi$.
• Hence, for transmitted light, minima occur when

$2\mu t = m\lambda$

where $m$ is an integer.

So, two successive minima correspond to a change in thickness

$\Delta t = \frac{\lambda}{2\mu}$

Given:

$\lambda = 560\text{ nm} = 560\times 10^{-9}\text{ m}$

Thus,

$\Delta t = \frac{560\times 10^{-9}}{2\times 1.4} = \frac{560\times 10^{-9}}{2.8} = 200\times 10^{-9} = 2\times 10^{-7}\text{ m}$

This change happens every $12$ s, so the rate of decrease of thickness is

$\frac{dt}{dt} = \frac{2\times 10^{-7}}{12} = \frac{1}{6}\times 10^{-7}\text{ m/s}$

2. Area of the circular film

Radius of ring:

$r=1.8\text{ cm}=1.8\times 10^{-2}\text{ m}$

Area:

$A=\pi r^2 = \pi(1.8\times 10^{-2})^2$

$A=\pi\times 3.24\times 10^{-4}\text{ m}^2$

3. Rate of evaporation (rate of decrease of volume)

Since the film is flat on both sides, volume is

$V=At$

Therefore,

$\frac{dV}{dt}=A\frac{dt}{dt}$

Substitute values:

$\frac{dV}{dt}=\pi\times 3.24\times 10^{-4}\times \frac{2\times 10^{-7}}{12}$

$\frac{dV}{dt}=\pi\times 3.24\times 10^{-4}\times \frac{1}{6}\times 10^{-7}$

$\frac{dV}{dt}=\pi\times 0.54\times 10^{-11}$

$\frac{dV}{dt}=5.4\pi\times 10^{-12}\text{ m}^3/\text{s}$

Writing this as

$\underline{\hspace{1cm}}\pi\times 10^{-13}\text{ m}^3/\text{s}$

we get

$5.4\pi\times 10^{-12} = 54\pi\times 10^{-13}$

So the required integer is:

$\boxed{54}$