Answer:
Option C
Explanation:
In the unit cell number of Cu atoms (fcc/ccp)
= $8\times\frac{1}{6}+6\times\frac{1}{2}=4$
As Ag atoms occupying edge centred = $12\times\frac{1}{4}=3$
and Au atoms are presents at the body centred=1
$\therefore$ formula , $Cu_{4}Ag_{3}Au$