Answer
$n_{i}=6$
Work Step by Step
The change in energy when transition occurs is equal to the energy of the photon emitted.
$\implies h\nu=2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$
$\implies (6.626\times10^{-34}\,J\cdot s)(114\times10^{12}\,Hz)=$
$2.18\times10^{-18}\,J(\frac{1}{4^{2}}-\frac{1}{n_{i}^{2}})$
$\implies 7.55364 \times10^{-20}\,J-\frac{2.18\times10^{-18}\,J}{16}=-\frac{2.18\times10^{-18}\,J}{n_{i}^{2}}$
$\implies -6.07136\times10^{-20}\,J=-\frac{2.18\times10^{-18}\,J}{n_{i}^{2}}$
Then,
$n_{i}^{2}=\frac{2.18\times10^{-18}\,J}{6.07136\times10^{-20}\,J}=36$
$\implies n_{i}=6$
