Given the voltage \( V = 4.2 \) volts and the battery capacity \( 5800 \) mAh, we can calculate the energy stored in the battery using the formula:
\[
\text{Energy supplied by battery} = Vq
\]
where \( q \) is the charge in coulombs. Converting \( 5800 \) mAh to coulombs:
\[
q = 5800 \times 3600 \times 10^{-3} \, \text{C} = 5800 \times 3.6 \, \text{C} = 20880 \, \text{C}
\]
Thus, the energy supplied by the battery is:
\[
\text{Energy} = 4.2 \times 5800 \times 3600 \times 10^{-3} = 87.696 \, \text{kJ}
\]
Therefore, the energy stored in the battery when fully charged is approximately \( 87.7 \, \text{kJ} \).