Answer
$\Delta_{120} = -91~MeV/c^2$
Work Step by Step
We can find the mass excess of $^{120} Sn$:
$\Delta_{120} = M - A$
$\Delta_{120} = 119.902 197~u - 120~u$
$\Delta_{120} = -0.097803~u$
We can express this mass excess in units of $MeV/c^2$:
$\Delta_{120} = -0.097803~u$
$\Delta_{120} = (-0.097803)~(1.66053886\times 10^{-27}~kg)(\frac{1~MeV}{1.6\times 10^{-13}~J})[\frac{(3.0\times 10^8~m/s)^2}{c^2}]$
$\Delta_{120} = -91~MeV/c^2$
