To determine which color of visible light will cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to consider the relationship between the photon energy, the work function, and the wavelength of light.

The maximum kinetic energy ($ \text{KE}_{\max} $) of the emitted photoelectrons is given by:

$ \text{KE}_{\max} = \frac{hc}{\lambda} - \phi $

For photoemission to occur, the energy of the incident photons must be greater than the work function ($\phi$):

$ \frac{hc}{\lambda} > \phi $

This implies:

$ \lambda < \frac{hc}{\phi} $

Substitute the values $ h = 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} $, $ c = 3 \times 10^{8} \, \text{m/s} $, and $\phi = 3 \, \text{eV}$:

$ \lambda < \frac{1242}{3} \, \text{nm} \approx 414 \, \text{nm} $

Since the wavelength must be less than 414 nm to cause emission, we need light with a shorter wavelength. Among the visible colors, blue light has a wavelength in this range, making it the correct choice. Thus, blue light will cause the emission of photoelectrons with this work function.