Step 1: Write down the given values: Wavelength of radiation, $\lambda = 256.7 \text{ nm} = 256.7 \times 10^{-9} \text{ m}$. Stopping voltage, $V_0 = 0.35 \text{ V}$. We use Planck's constant $h = 6.626 \times 10^{-34} \text{ J s}$, speed of light $c = 3 \times 10^8 \text{ m s}^{-1}$, and charge of electron $e = 1.602 \times 10^{-19} \text{ C}$.

Step 2: State the photoelectric effect equation: $h\nu = W_0 + KE$. The kinetic energy of the ejected electron is equal to the stopping potential energy, so $KE = eV_0$. Therefore, the work function is $W_0 = h\nu - eV_0 = \frac{hc}{\lambda} - eV_0$.

Step 3: Calculate the energy of the incident photon: $h\nu = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m s}^{-1})}{256.7 \times 10^{-9} \text{ m}} = 7.743 \times 10^{-19} \text{ J}$. Converting to eV (since $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$): $h\nu = \frac{7.743 \times 10^{-19}}{1.602 \times 10^{-19}} = 4.83 \text{ eV}$.

Step 4: Calculate the kinetic energy from the stopping voltage: $KE = eV_0 = 0.35 \text{ eV}$.

Step 5: Calculate the work function for silver metal: $W_0 = h\nu - KE = 4.83 \text{ eV} - 0.35 \text{ eV} = 4.48 \text{ eV}$. In Joules, $W_0 = 4.48 \times 1.602 \times 10^{-19} \text{ J} = 7.18 \times 10^{-19} \text{ J}$.