1. Energy levels of hydrogen atom

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

Also, for a Coulomb bound system, $U_n = 2E_n$ because total energy is half the potential energy.

2. Initial state: first excited state

The first excited state corresponds to $n=2$ So the initial energy is $E_i = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4\text{ eV}$

3. Final state from given potential energy

Given final potential energy of electron: $U_f = -1.08\text{ eV}$

Using $U_f = 2E_f$ we get $E_f = \frac{U_f}{2} = \frac{-1.08}{2} = -0.54\text{ eV}$

4. Photon energy absorbed

The absorbed photon energy is the difference in total energies: $\Delta E = E_f - E_i$ $\Delta E = (-0.54) - (-3.4) = 2.86\text{ eV}$

The question says photon energy is $x \times 10^{-2}\text{ eV}$ So, $x \times 10^{-2} = 2.86$

Thus, $x = 2.86 \times 10^2 = 286$

5. Final answer

$\boxed{286}$