Answer
${\bf 8.4}\, \text{MeV}$
Note: This is a rough estimate based on the graph, so your solution may vary slightly from ours.
Work Step by Step
First, we need to calculate the Binding Energy of a Single $ ^4\text{He} $ Nucleus.
From the mentioned graph, the binding energy per nucleon for a $ ^4\text{He} $ nucleus is approximately $ 7.0\, \text{MeV} $.
Since a $ ^4\text{He} $ nucleus has a mass number $ A = 4 $, the total binding energy $ B_{\text{He}} $ for one $ ^4\text{He} $ nucleus is:
$$
B_{\text{He}} = \text{Binding energy per nucleon} \times \text{Mass number}
$$
Substitute the values:
$$
B_{\text{He}} = 7 \, \text{MeV} \times 4 = \bf 28 \, \rm {MeV}
$$
So, the Total Binding Energy of Three $ ^4\text{He} $ Nuclei is
$$
B_{3\text{He}} = 3 \times B_{\text{He}}=3(28)=\bf 84\;\rm MeV
$$
Now we need to calculate the Binding Energy of the $ ^{12}\text{C} $ Nucleus. The binding energy per nucleon for $ ^{12}\text{C} $ (carbon-12) from the graph is approximately $ 7.7 \, \text{MeV} $.
Since $ ^{12}\text{C} $ has a mass number $ A = 12 $, the total binding energy $ B_{\text{C}} $ of the $ ^{12}\text{C} $ nucleus is:
$$
B_{\text{C}} = \text{Binding energy per nucleon} \times \text{Mass number} \\
= 7.7 \, \text{MeV} \times 12 = \bf 92.4 \, \rm {MeV}
$$
To Calculate the Total Energy Released During Fusion which is the difference between the binding energy of the $ ^{12}\text{C} $ nucleus and the combined binding energy of the three $ ^4\text{He} $ nuclei:
$$
E_{\text{released}} = B_{\text{C}} - B_{3\text{He}}
$$
Substitute the values:
$$
E_{\text{released}} = 92.4 \, \text{MeV} - 84\, \text{MeV}
$$
$$
E_{\text{released}} =\color{red} {\bf 8.4}\, \text{MeV}
$$
