If a proton and an anti-proton come close to each other and annihilate, how much energy will be released?

In the nuclear process $n \to p + e^- + \bar{\nu}$,if the masses of proton,neutron,and electron are $1.6725 \times 10^{-27} \ kg$,$1.6747 \times 10^{-27} \ kg$,and $9 \times 10^{-31} \ kg$ respectively,then the energy released is ...... $MeV$.

From the given data,the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} {Al}$ is $x \times 10^{-3} {J}$.
Mass of neutron $= 1.00866 \, {u}$
Mass of proton $= 1.00726 \, {u}$
Mass of aluminium nucleus $= 26.98154 \, {u}$
(Assume $1 \, {u}$ corresponds to $1 \, {J}$ of energy for the purpose of this calculation)
(Round off to the nearest integer)

$M_n$ and $M_p$ represent the mass of a neutron and a proton, respectively. If an element having atomic mass $M$ has $N$ neutrons and $Z$ protons, then the correct relation is:

Mass numbers of the elements $A, B, C$ and $D$ are $30, 60, 90$ and $120$ respectively. The specific binding energies of them are $5 \text{ MeV}, 8.5 \text{ MeV}, 8 \text{ MeV}$ and $7 \text{ MeV}$ respectively. In which of the following reaction$(s)$ is energy released?
$1. D \rightarrow 2B$
$2. C \rightarrow B + A$
$3. B \rightarrow 2A$

$A$ nucleus $^{A}_{Z} X$ has mass represented by $M(A, Z)$. If $M_p$ and $M_n$ denote the mass of a proton and a neutron respectively, and $B.E.$ is the binding energy in $MeV$, then: