hello Pratiksha,

The given radius of FCC XENON Crystal is 620 pm.then the atomic radius of Xe is  found by.

For FCC Crystal,

Edge=2*R*underroot of 2.

Edge=620pm........given

620=2*R*underroot of 2

620/2=R*1.414.............underoot 2=1.414

310=R*1.414

310/1.414=R

R=219.86 pm

hence atomic radius of xe is 219.86.

hope this may help you