A particular hydrogen - like ion emits the radiation of frequency 3...

For Hydrogen like atom, energy emitted is $E = - 13 . 6{Z}^{2}\left(1{n}_{i}^{2} - 1{n}_{f}^{2}\right)$

$⇒ E = C\left(1{n}_{f}^{2} - 1{n}_{i}^{2}\right)$

$⇒ hν = C\left[1{n}_{f}^{2} - 1{n}_{i}^{2}\right]$

Therefore, the  ratio$\dfrac{{v}_{1}}{{v}_{2}} = \dfrac{{\left[1{n}_{f}^{2} - 1{n}_{i}^{2}\right]}_{2 - 1}}{{\left[1{n}_{f}^{2} - 1{n}_{i}^{2}\right]}_{3 - 1}}$

$= \dfrac{\left[\dfrac{1}{1} - \dfrac{1}{4}\right]}{\left[\dfrac{1}{1} - \dfrac{1}{9}\right]} = \dfrac{\dfrac{3}{4}}{\dfrac{8}{9}}$

$= \dfrac{3}{4} \times \dfrac{9}{8}$

$⇒ \dfrac{{v}_{1}}{{v}_{2}} = \dfrac{27}{32}$

$⇒ {v}_{2} = \dfrac{32}{27}{v}_{1} = \dfrac{32}{27} \times 3 \times {10}^{15}Hz = \dfrac{32}{9} \times {10}^{15}Hz$

Hence, $x = 32$.