Answer
The required weight of fissionable material is $~~24.5~N$
Work Step by Step
We can find the required mass of fissionable material:
$(\frac{0.080}{100})~mc^2 = 1.80\times 10^{14}~J$
$m = \frac{1.80\times 10^{16}~J}{0.080~c^2}$
$m = \frac{1.80\times 10^{16}~J}{(0.080)~(3.0\times 10^8~m/s)^2}$
$m = 2.5~kg$
We can find the weight:
$weight = mg = (2.5~kg)(9.8~m/s^2) = 24.5~N$
The required weight of fissionable material is $~~24.5~N$
