Answer
$\Delta_1 = 7.3~MeV/c^2$
Work Step by Step
We can find the mass excess of $^1 H$:
$\Delta_1 = M - A$
$\Delta_1 = 1.007825~u - 1~u$
$\Delta_1 = 0.007825~u$
We can express this mass excess in units of $MeV/c^2$:
$\Delta_1 = 0.007825~u$
$\Delta_1 = (0.007825)~(1.66053886\times 10^{-27}~kg)(\frac{1~MeV}{1.6\times 10^{-13}~J})[\frac{(3.0\times 10^8~m/s)^2}{c^2}]$
$\Delta_1 = 7.3~MeV/c^2$
