Write the nuclear reaction

Three helium nuclei (alpha particles) combine to form a carbon nucleus: $3\,{}^4_2\text{He} \rightarrow {}^{12}_6\text{C} + \text{energy}$

Use atomic masses

Given atomic mass of helium: $m(\text{He}) = 4.002603\,\text{u}$

For carbon, the atomic mass of ${}^{12}\text{C}$ is exactly: $m(^{12}\text{C}) = 12.000000\,\text{u}$

Since atomic masses are used on both sides, the electron masses cancel properly.

Find mass of reactants

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

Find mass defect

$\Delta m = m_{\text{initial}} - m_{\text{final}}$ $\Delta m = 12.007809 - 12.000000 = 0.007809\,\text{u}$

Convert mass defect into energy

Using: $1\,\text{u} = 931\,\text{MeV}/c^2$

So energy released is: $Q = \Delta m \times 931$ $Q = 0.007809 \times 931\,\text{MeV}$ $Q = 7.271179\,\text{MeV}$

Express in the required form

We need: $Q = \boxed{\_\_\_} \times 10^{-2}\,\text{MeV}$

Since: $7.271179\,\text{MeV} = 727.1179 \times 10^{-2}\,\text{MeV}$

Therefore, the required integer is: $\boxed{727}$

Comparison with stored answer

Derived answer = $727$

Stored correct answer = $727$

Hence, they agree.