Answer
$E_2 = 109~eV$
Work Step by Step
If the electron can absorb at the wavelength $\lambda_c$ or less, then the energy associated with the wavelength $\lambda_c$ will give the electron enough energy to reach a total of $450.0~eV$
We can find the energy associated with $\lambda_c = 2.9108~nm$:
$E = \frac{hc}{\lambda_c}$
$E = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{2.9108\times 10^{-9}~m}$
$E = 426.3~eV$
We can find the energy when the electron is in the ground state:
$E_1 = 450.0~eV-426.3~eV = 23.7~eV$
When the electron in the ground state absorbs at the longest wavelength, the electron jumps from $n=1$ to $n=2$
We can find the energy associated with $\lambda_a = 14.588~nm$:
$E = \frac{hc}{\lambda_a}$
$E = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{14.588\times 10^{-9}~m}$
$E = 85.06~eV$
We can find the energy in the first excited state where $n=2$:
$E_2 = E_1+85.06~eV$
$E_2 = 23.7~eV+85.06~eV$
$E_2 = 109~eV$
