Correct Answer - A
`rho=(ZxxM)/(a^3xxN_0xx10^(-30))`
or `a^3=(ZxxM)/(rhoxxN_0xx10^(-30))`
For ccp, i.e., f.c.c., Z=4.
Hence,`a^3=(4xx197)/(19.4xx6.02xx10^23xx10^(-30))`
`=6.747xx10^7`
=`67.47xx10^6`
or `a=(67.47)^(1//3)xx10^2`
`=4.07xx10^2` pm
=407 pm