Answer
$E \approx 26.4 MeV$
Work Step by Step
According to uncertainty principle,
$p \approx \Delta p \approx \frac{\Delta h}{\Delta x} \approx \frac{h}{r}$, So
$E = \frac{p^2}{2m} \approx \frac{(hc)^2}{2 (mc^2) r^2}$
$E = \frac{(1240 MeV.fm)^2}{2 (938 MeV) [1.2 fm [100)^{1/3}]^2}$
$E \approx 26.4 MeV$
