1. Potential energy in hydrogen atom

For an electron in the $n^{\text{th}}$ Bohr orbit of hydrogen,

$E_n = -\frac{13.6}{n^2}\,\text{eV}$

Also, for a Coulomb-bound system,

$U_n = V_n = 2E_n = -\frac{27.2}{n^2}\,\text{eV}$

So the potential energy varies as

$V_n \propto -\frac{1}{n^2}$

2. Use the given ratio

Given,

$\frac{V_i}{V_f} = 6.25 = \frac{25}{4}$

Now,

$V_i = -\frac{27.2}{n_i^2}, \qquad V_f = -\frac{27.2}{n_f^2}$

Therefore,

$\frac{V_i}{V_f} = \frac{-27.2/n_i^2}{-27.2/n_f^2} = \frac{n_f^2}{n_i^2}$

Hence,

$\frac{n_f^2}{n_i^2} = \frac{25}{4}$

Taking square root,

$\frac{n_f}{n_i} = \frac{5}{2}$

So,

$n_f = \frac{5}{2} n_i$

3. Find the smallest possible integer value of $n_f$

Since $n_i$ and $n_f$ must both be positive integers, $n_i$ must be even.

The smallest possible even value is

$n_i = 2$

Then,

$n_f = \frac{5}{2}\times 2 = 5$

4. Final answer

The smallest possible value of $n_f$ is

$\boxed{5}$