Explanation:
Binding energy per nucleon
For \(_1{H^2},\,\,\,\,\,{E_1} = \frac{{2.22}}{2} = 1.11MeV\)
For \({\,_2}H{e^4},\,\,\,\,\,{E_2} = \frac{{28.3}}{4} = 7.08MeV\)
For \({\,_{28}}F{e^{56}},\,\,\,\,\,{E_3} = \frac{{492}}{{56}} = 8.80MeV\)
For \({\,_{92}}{U^{235}},\,\,\,\,\,{E_4} = \frac{{786}}{{235}} = 3.35MeV\)
Hence, it is obvious that \({}_{28}F{e^{56}}\) is most stable.