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 :

The stability of a nucleus is measured by its total binding energy: larger binding energy ⇒ greater stability.
Let the binding energies of $${}_{2}He^{3},\; {}_{2}He^{4}$$ and $${}_{2}He^{5}$$ be $$B_3,\; B_4$$ and $$B_5$$ (in MeV).

Write the given reactions in the form
$${}_{2}He^{3} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{4} + 20\;{\rm MeV}$$
$${}_{2}He^{4} + {}_{0}n^{1} \;\longrightarrow\; {}_{2}He^{5} - 0.9\;{\rm MeV}$$

For any nuclear reaction,
energy released $$=\;(\text{binding energy of products})-(\text{binding energy of reactants})$$.

Case 1:
Products’ binding energy $$=\;B_4$$, reactants’ binding energy $$=\;B_3 + 0$$ (a free neutron has zero binding energy).
Hence
$$B_4 - (B_3 + 0) = 20$$ $$-(1)$$
$$\Rightarrow\; B_4 = B_3 + 20$$

Case 2:
Products’ binding energy $$=\;B_5$$, reactants’ binding energy $$=\;B_4 + 0$$.
Here the reaction absorbs 0.9 MeV, so energy released = $$-0.9$$ MeV:
$$B_5 - (B_4 + 0) = -0.9$$ $$-(2)$$
$$\Rightarrow\; B_5 = B_4 - 0.9$$

Substitute $$B_4$$ from $$(1)$$ into $$(2)$$:
$$B_5 = (B_3 + 20) - 0.9 = B_3 + 19.1$$

Thus
$$B_4 = B_3 + 20$$
$$B_5 = B_3 + 19.1$$
Since $$20 \gt 19.1 \gt 0$$, we get the order
$$B_4 \gt B_5 \gt B_3$$.

Greater binding energy means higher stability, so
$$X_4 \gt X_5 \gt X_3$$.

Therefore the correct option is Option A.