An electron in a hydrogen atom is initially in an excited state...

For an electron in a hydrogen-like atom, the potential energy (PE) is related to the kinetic energy (KE) by $\text{PE} = -2\text{KE}$.

In the initial excited state, the kinetic energy is given as $1.51\text{ eV}$.
Initial potential energy, $\text{PE}_{i} = -2 \times 1.51\text{ eV} = -3.02\text{ eV}$.

In the ground state of a hydrogen atom, the total energy (TE) is $-13.6\text{ eV}$.
The potential energy is related to the total energy by $\text{PE} = 2\text{TE}$.
Final potential energy, $\text{PE}_{f} = 2 \times (-13.6\text{ eV}) = -27.2\text{ eV}$.

The change in electrostatic potential energy is:
$\Delta\text{PE} = \text{PE}_{f} - \text{PE}_{i}$
$\Delta\text{PE} = -27.2\text{ eV} - (-3.02\text{ eV}) = -24.18\text{ eV}$.

Answer: $-24.18\text{ eV}$