What is the order of binding energy per nucleon?

The binding energy for the following nuclear reactions are expressed in $MeV$.
${ }_2 He ^3+{ }_0 n ^1 \rightarrow{ }_2 He ^4+20 \ MeV$
${ }_2 He ^4+{ }_0 n ^1 \rightarrow{ }_2 He ^5-0.9 \ MeV$
If $X_3, X_4, X_5$ denote the stability of ${ }_2 He ^3, { }_2 He ^4$ and ${ }_2 He ^5$, respectively, then the correct order is:

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. The speed of light is $c$. The binding energy per nucleon for the parent nucleus is $E_1$ and that for the daughter nuclei is $E_2$. Then:

The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$
The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is:

The mass of a proton is $1.0073 \; u$ and that of a neutron is $1.0087 \; u$ ($u =$ atomic mass unit). The binding energy of ${ }_2^4 \text{He}$ is (Given: helium nucleus mass $\approx 4.0015 \; u$):

The masses of a proton, neutron, and helium nucleus are $1.0073\,u$, $1.0087\,u$, and $4.0015\,u$ respectively. The binding energy of the helium nucleus is $.........\,MeV$.