To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein's equation $$E = \Delta mc^2$$, where $$E$$ is the energy released, $$\Delta m$$ is the mass defect, and $$c$$ is the speed of light.

The atomic mass of a helium nucleus (also known as an alpha particle) is $$4.002603 \, \text{u}$$.

1. Calculate the total initial mass of three helium nuclei:

$$\text{Total initial mass} = 3 \times 4.002603 \, \text{u} = 12.007809 \, \text{u}$$

2. The atomic mass of a carbon nucleus formed by the fusion of three helium nuclei is not directly given, but we can infer it's approximately $$12 \, \text{u}$$, based on knowledge of isotopes and considering that the question appears to simplify the carbon nucleus to a mass number of 12 (common carbon-12 isotope).

3. Calculate the mass defect ($$\Delta m$$):

$$\Delta m = \text{Total initial mass} - \text{Final mass}$$

$$\Delta m = 12.007809 \, \text{u} - 12 \, \text{u} = 0.007809 \, \text{u}$$

4. Convert the mass defect to energy. Given $$1 \, \text{u} = 931 \, \text{MeV/c}^2$$, the energy released is calculated using the formula $$E = \Delta mc^2$$:

$$E = 0.007809 \, \text{u} \times 931 \, \text{MeV/u} = 7.271839 \, \text{MeV}$$

Since the question asks for the answer in the format of $$\times 10^{-2} \, \text{MeV}$$, we convert the energy released:

$$7.271839 \, \text{MeV} = 727.1839 \times 10^{-2} \, \text{MeV}$$

Therefore, the energy released in this reaction is approximately $$727.1839 \times 10^{-2} \, \text{MeV}$$. The exact value might differ slightly depending on how the atomic mass of the carbon nucleus is considered or rounded in specific scenarios, but based on the information provided, this is a suitable approximation.