Answer
No.
Work Step by Step
From the mentioned graph, the binding energy per nucleon for $ ^{56}\text{Fe} $ is about $ 8.8 \, \text{MeV} $.
So, the total binding energy $ B_{\text{Fe}} $ for the $ ^{56}\text{Fe} $ nucleus is:
$$
B_{\text{Fe}} = \text{Binding energy per nucleon} \times \text{Mass number}
$$
Substitute the values:
$$
B_{\text{Fe}} = 8.8 \, \text{MeV} \times 56 = 493 \, \text{MeV}
$$
The binding energy per nucleon for $ ^{28}\text{Al} $ is about $ 8.4 \, \text{MeV} $ from the graph.
So, the total binding energy $ B_{\text{Al}} $ for a single $ ^{28}\text{Al} $ nucleus is:
$$
B_{\text{Al}} = \text{Binding energy per nucleon} \times \text{Mass number}
$$
Substitute the values:
$$
B_{\text{Al}} = 8.4 \, \text{MeV} \times 28 = 235 \, \text{MeV}
$$
And hence,
$$
B_{\text{2Al}} = 2 \times B_{\text{Al}} = 2 \times 235 \, \text{MeV} = \bf 470 \rm \, {MeV}
$$
Now we need to Compare the Binding Energies.
The total binding energy of the original $ ^{56}\text{Fe} $ nucleus is $ 493 \, \text{MeV} $, which is higher than the combined binding energy of the two$\; ^{28}\text{Al} $ nuclei.
$$
(B_{\text{Fe}} = 493 \, \text{MeV}) > (B_{\text{2Al}} = 470 \, \text{MeV})
$$
Since the binding energy of the $ ^{56}\text{Fe} $ nucleus is greater, this means the $ ^{56}\text{Fe} $ nucleus is more tightly bound than the two $ ^{28}\text{Al} $ nuclei.
So, to force $ ^{56}\text{Fe} $ to fission into two $ ^{28}\text{Al} $ nuclei, an energy input is required.
Therefore, the $ ^{56}\text{Fe} $ nucleus$\underline{\color{red}{\text{ cannot}}}$ spontaneously fission into two $ ^{28}\text{Al} $ nuclei because its binding energy is higher, indicating it is more stable.
