Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 51

$1.2\times 10^6Kg$

We can find the required mass as follows: $N=\frac{E}{E_U}$ $N=\frac{8.42\times 10^9J}{173MeV}(\frac{1MeV}{1.6\times 10^{-13}})=3.035\times 10^{30}$ Now we can find the mass $m=Nm_u$ We plug in the known values to obtain: $m=3.035\times 10^{30}(235.0u)(1.66\times 10^{-27}Kg/u)$ $m=1.2\times 10^6Kg$