Step 1: Convert the mass defect from atomic mass units (a.m.u.) to kilograms (kg). Given mass defect ($\Delta m_{amu}$) = $0.05 \, \text{a.m.u.}$ Given conversion factor: $1 \, \text{a.m.u.} = 1.67 \times 10^{-27} \, \text{kg}$ The mass defect in kilograms ($\Delta m$) is: $$\Delta m = 0.05 \, \text{a.m.u.} \times (1.67 \times 10^{-27} \, \text{kg/a.m.u.})$$ $$\Delta m = 0.0835 \times 10^{-27} \, \text{kg}$$ $$\Delta m = 8.35 \times 10^{-29} \, \text{kg}$$ Step 2: Use Einstein's mass-energy equivalence formula to calculate the energy released. The formula is $E = \Delta m c^2$, where: $E$ is the energy released $\Delta m$ is the mass defect in kilograms $c$ is the speed of light in a vacuum Given speed of light ($c$) = $3.0 \times 10^8 \, \text{m/s}$ Substitute the values into the formula: $$E = (8.35 \times 10^{-29} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2$$ $$E = (8.35 \times 10^{-29}) \times (9.0 \times 10^{16}) \, \text{J}$$ $$E = (8.35 \times 9.0) \times (10^{-29} \times 10^{16}) \, \text{J}$$ $$E = 75.15 \times 10^{(-29+16)} \, \text{J}$$ $$E = 75.15 \times 10^{-13} \, \text{J}$$ Step 3: Express the energy in standard scientific notation. $$E = 7.515 \times 10^{-12} \, \text{J}$$ The amount of energy released is $\boxed{\text{7.515 \times 10^{-12} J}}$.