Given mass defect ( m_amu) = 0.05 a.m.u.

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:

$$\Delta m = 0.05 \, \text{a.m.u.} \times (1.67 \times 10^{-27} \, \text{kg/a.m.u.})$$ $$\Delta m = 0.0835 \times 10^{-27} \, \text{kg}$$ $$\Delta m = 8.35 \times 10^{-29} \, \text{kg}$$

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:

$$E = (8.35 \times 10^{-29} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2$$ $$E = (8.35 \times 10^{-29}) \times (9.0 \times 10^{16}) \, \text{J}$$ $$E = (8.35 \times 9.0) \times (10^{-29} \times 10^{16}) \, \text{J}$$ $$E = 75.15 \times 10^{(-29+16)} \, \text{J}$$ $$E = 75.15 \times 10^{-13} \, \text{J}$$

$$E = 7.515 \times 10^{-12} \, \text{J}$$

The amount of energy released is \boxed{\text{7.515 \times 10^{-12} J}}.