Which one of the following electrons in the ground state will have the least amount of energy?

The energy of an electron in an atom is negative, with the value zero corresponding to a free electron at rest. 'Least energy' implies the most negative energy value, corresponding to the most stable and tightly bound state.

1. Hydrogen (1s): The electron occupies the $n=1$ shell. The energy of the electron in the ground state of Hydrogen is $E_1 = -2.18 \times 10^{-18} \text{ J}$ (or $-13.6 \text{ eV}$).
2. Carbon (2p): The specified electron is in the $2p$ orbital ($n=2$). Although the nuclear charge ($Z=6$) is higher, the shielding by inner $1s$ electrons and the higher principal quantum number raise its energy compared to H($1s$). The first ionization enthalpy of Carbon is roughly $1086 \text{ kJ mol}^{-1}$ ($E \approx -11.3 \text{ eV}$), which is higher (less negative) than Hydrogen.
3. Sodium (3s): The outermost electron is in the $3s$ orbital ($n=3$). Due to shielding by inner shells ($n=1, 2$) and the higher $n$, it is loosely bound. Its ionization enthalpy is $496 \text{ kJ mol}^{-1}$ ($E \approx -5.1 \text{ eV}$).
4. Copper (4s): The specified electron is in the $4s$ orbital ($n=4$). Its ionization enthalpy is $745 \text{ kJ mol}^{-1}$ ($E \approx -7.7 \text{ eV}$).

Comparing the energy levels, the electron in the Hydrogen atom ($n=1$) has the most negative energy, thus the least amount of energy.