1. Use Einstein’s photoelectric equation

For incident radiation of wavelength $\lambda$, the stopping potential $V_s$ satisfies

$eV_s = \frac{hc}{\lambda} - \phi$

where $\phi$ is the work function.

Let the threshold wavelength be $\lambda_0 = n\lambda$. Then

$\phi = \frac{hc}{\lambda_0} = \frac{hc}{n\lambda}$

2. Write equation for wavelength $\lambda$

Given stopping potential is $V$:

$eV = \frac{hc}{\lambda} - \frac{hc}{n\lambda}$

$eV = \frac{hc}{\lambda}\left(1 - \frac{1}{n}\right) \quad ...(1)$

3. Write equation for wavelength $3\lambda$

Given stopping potential is $\dfrac{V}{4}$:

$e\left(\frac{V}{4}\right) = \frac{hc}{3\lambda} - \frac{hc}{n\lambda}$

$\frac{eV}{4} = \frac{hc}{\lambda}\left(\frac{1}{3} - \frac{1}{n}\right) \quad ...(2)$

4. Divide (2) by (1)

From (1) and (2):

$\frac{\frac{eV}{4}}{eV} = \frac{\frac{1}{3} - \frac{1}{n}}{1 - \frac{1}{n}}$

$\frac{1}{4} = \frac{\frac{1}{3} - \frac{1}{n}}{1 - \frac{1}{n}}$

Now simplify:

$\frac{1}{4} = \frac{\frac{n-3}{3n}}{\frac{n-1}{n}} = \frac{n-3}{3(n-1)}$

So,

$\frac{1}{4} = \frac{n-3}{3(n-1)}$

Cross-multiplying:

$3(n-1) = 4(n-3)$

$3n - 3 = 4n - 12$

$n = 9$

5. Final answer

$\boxed{9}$

The derived answer matches the stored correct answer.