According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi$.

Case 1: For wavelength $\lambda$ and stopping potential $3V_0$: $e(3V_0) = \frac{hc}{\lambda} - \phi$ ...(i)

Case 2: For wavelength $2\lambda$ and stopping potential $V_0$: $e(V_0) = \frac{hc}{2\lambda} - \phi$ ...(ii)

Substituting $eV_0$ from (ii) into (i): $3(\frac{hc}{2\lambda} - \phi) = \frac{hc}{\lambda} - \phi$ $\frac{3hc}{2\lambda} - 3\phi = \frac{hc}{\lambda} - \phi$ $\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = 2\phi$ $\frac{hc}{2\lambda} = 2\phi \implies \phi = \frac{hc}{4\lambda}$

The threshold wavelength $\lambda_0$ is related to the work function by $\phi = \frac{hc}{\lambda_0}$. Therefore, $\frac{hc}{\lambda_0} = \frac{hc}{4\lambda} \implies \lambda_0 = 4\lambda$.