Why dust is optically thin in Far Infrared wavelengths? What is the actual meaning of the statement 'Dust is optically thin in the Far Infrared (FIR) over most of the Galaxy'? Kindly Help
The term "optically thin" means that the optical depth is small. The optical depth is a measure of the opacity of a medium, in this case dust, experienced by light traveling through that medium, and is defined as
$$
\tau \equiv n \, r \, \sigma,
$$
where $n$ is the density of the particles in question, $r$ is the distance traveled through the medium, and $\sigma = \sigma(\lambda)$ is the cross section of the particles, which is dependent on the wavelength $\lambda$ of the photons. If $\tau\ll1$, the medium is said to be optically thin, while if $\tau\gg1$, it is said to be optically thick. The fraction of the light that is extinguished by the journey through the medium is $e^{-\tau}$, so the two cases indicate when the light is mostly transferred and mostly extinguished, respectively.

Cosmic dust is composed of particles spanning a large range of sizes. Photons interacting with the dust are either scattered in another direction, or absorbed, depending on the albedo of the dust. But in both cases, a photon has a larger probability of interacting with a dust grain which is comparable to its wavelength. Because of the size distribution, an ensemble of dust grains hence has a characteristic extinction curve. In the figure below (modified from Laursen et al. 2009), I plotted the (functional fits to the observed) extinction curves of dust in the Small (dashed line) and Large (solid line) Magellanic Clouds$^\dagger$:



The extinction is here given in terms of "cross section per hydrogen atom", but you can just think of it as an average dust grain cross section. The colors show the ultraviolet (purple) and infrared (red) regions of the spectrum.
You can see that the extinction, or the cross section, is largest for UV photons, and as you go to longer wavelengths, the cross section decreases sharply and is very small in the far infrared ($\lambda \gtrsim 30 \, \mu\mathrm{m}$; the exact definitions depend on your field of interest).

This is what your statement refers to. But in fact it's badly phrased, because the optical depth is also a function of density and distance, so for instance for a distance $r = 1 \, \mathrm{cm}$ through a dust cloud, the optical depth is $\ll1$ for all photons.



$^\dagger$Extinction curves in other galaxies look similar. In particular, the extinction along most sightlines in the Milky Way features the same bump around $\lambda \simeq 2175$ Å. The origin of this bump still isn't well understood, but may be partly due to graphites and/or PAHs.