Answer
See explanation.
Work Step by Step
We calculate the binding energy of $_{\space \space 6}^{12}C$
$E_B = (Zm_H + (A-Z)m_n -m_{C-11})c^2$
$E_B = ((6)(1.007825 u) + (12-6)(1.008665 u) - 12.000000 u)931.5 MeV/u$
$E_B =92.16 MeV$
Now, for the $alpha$ particle, $_{ 2}^{4}H$
$E_B = (Zm_H + (A-Z)m_n -m_{C-11})c^2$
$E_B = (2(1.007825 u) + (4-2)(1.008665 u) – 4.002603 u))(931.5 MeV/u) $
$E_B = 28.296 MeV$
We can conclude that the binding energy of $_{\space \space 6}^{12}C$ is 3 times greater than the binding energy of an alpha particle. This is due to the fact that it takes three times more energy to bind the three alpha particles in the nucleus in order to form a carbon atom.
