Are you talking about imaging the nanoparticle? It is way below the resolution limit.

Hugh: in my understanding the question is about the size of the image produced by the nanoparticle.

The resolution limit d of the microscope at 610 nm is lambda/2*NA, i.e. 610 nm / 1.8 = 339 nm (Abbe criterion). This gives a circular area of the diffraction limited image formed by the nanoparticle of pi*d*d/4 = 0.7853 * 0.339 µm * 0.339 µm = 0.09 µm2.

Not sure what is meant with 'magnified diffraction limited image', but I guess you have to magnify the diameter d by 30, which is the magnification of the microscope. In this case d would be 30*339nm = 10.166 µm and the corresponding area is 81.17 µm2 (no warranty).

I agree with the answer given by Peter Lash. My only addition is that the 30X magnification implies that that objective is mounted in a microscope with the corresponding tube lens. Different manufacturers use different tube lens focal lengths (160-250mm) despite of the fact that yo may easily adapt objectives of various brands into pretty much any microscope. Often times in single nanoparticle work you build your own setup using arrangements very different from a commercial microscope. Then it is relevant to correct for a different focal length of the "tube lens" . If you have additional lenses carrying the emission into the detector the area of the image may also change.

But the 10.166 micrometers is the answer for an optical system composed of a 30x objective and a tube lens corresponding to that objective according to the manufacturer.

I agree Peter, but the image is of the fluorescence, and so it depends on the brightness, and the sensitivity of the detector. If it is very weak, the detector will not detect it. If it is very bright, it can saturate, and appear much larger in the image.