Answer
$E = 7.28~MeV$
Work Step by Step
We can find the "extra" mass when the three alpha particles fuse:
$m = (3)(4.0026~u)-12.0000~u$
$m = (0.0078)~u$
$m = (0.0078)~(1.66\times 10^{-27}~kg)$
$m = 1.2948\times 10^{-29}~kg$
This "extra" mass is transformed into energy.
We can find the energy that is released:
$E = mc^2$
$E = (1.2948\times 10^{-29}~kg)(3.0\times 10^8~m/s)^2$
$E = 1.16532\times 10^{-12}~J$
$E = (1.16532\times 10^{-12}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 7.28\times 10^6~eV$
$E = 7.28~MeV$
