Use the hydrogen energy-level formula

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

The energy of the emitted photon in a transition from level $n_i$ to $n_f$ is $\Delta E = 13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ where $n_i > n_f$.

Photon energy for transition from third level to second level

Here $n_i=3$ and $n_f=2$.

So, $E_1 = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ $E_1 = 13.6\left(\frac{1}{4}-\frac{1}{9}\right)$ $E_1 = 13.6\left(\frac{9-4}{36}\right)$ $E_1 = 13.6\cdot \frac{5}{36}$

Photon energy for transition from highest permitted level to second level

The highest permitted level means $n_i=\infty$.

So, $E_2 = 13.6\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)$ Since $\frac{1}{\infty^2}=0$, $E_2 = 13.6\cdot \frac{1}{4}$

Form the given ratio

Given ratio is $\frac{E_1}{E_2} = \frac{x}{x+4}$

Now, $\frac{E_1}{E_2} = \frac{13.6\cdot \frac{5}{36}}{13.6\cdot \frac{1}{4}}$ $\frac{E_1}{E_2} = \frac{5}{36}\cdot 4 = \frac{5}{9}$

Hence, $\frac{x}{x+4} = \frac{5}{9}$

Solve for $x$

$9x = 5(x+4)$ $9x = 5x + 20$ $4x = 20$ $x = 5$

Comparison with stored answer

Derived answer = $5$

Stored correct answer = $5$

Both match.