We can solve the exercise by using Heisenberg's principle. In its energy-time version, Heisenberg principle states that the product between the uncertainty on the energy and on the time is larger than:
[tex]\Delta E \Delta t \ \textgreater \ \frac{h}{4 \pi} [/tex] (1)
where [tex]\Delta E, \Delta t[/tex] are the uncertainties on the energy and on the time, and h is the Planck constant.
The lifetime of the particle is 2.05 ms, so we can assume the maximum uncertainty on the time corresponds to the lifetime itself:
[tex]\Delta t = 2.05 ms = 2.05 \cdot 10^{-3} s[/tex]
And so the minimum uncertainty on the energy can be found by using (1):
[tex]\Delta E \ \textgreater \ \frac{h}{4 \pi \Delta t}= \frac{6.6 \cdot 10^{-34} Js}{4 \pi (2.05 \cdot 10^{-3} s)}= 2.56 \cdot 10^{-32} J[/tex]
Keeping in mind that [tex]1 eV = 1.6 \cdot 10^{-19}J[/tex], we can convert the energy uncertainty into electronvolts:
[tex]\Delta E = 2.56 \cdot 10^{-32} J : 1,6 \cdot 10^{-19} J/eV = 1,6 \cdot 10^{-13} eV[/tex]
