The limit of resolution of a microscope is given by \(x =\frac{0.61 \lambda}{\mu \sin \theta}\)
It is given that \(\lambda=6 \times 10^{-7} m\), and the numerical aperture \(\mu \sin \theta=0.12\). Therefore,
\(x =\frac{0.61 \times 6 \times 10^{-7}}{0.12}=3.05 \times 10^{-6} m \approx 3 \mu m\)