Density of fcc = $\large\frac{Z_1 \times At.mass}{Av.no\times V_1}$
and density in bcc =$ \large\frac{Z_2\times At.mass}{Av.no.\times V_2}$
$\large\frac{d_{fcc}}{d_{bcc}} = \large\frac{Z_1}{Z_2}\times \large\frac{V_2}{V_1}$
For fcc
$Z_1 = 4 ; V_1 = a^3 = (3.5\times10^{-8})^3$
For bcc
$Z_2 = 2 ; V_2 = a^3 = (3.0\times10^{-8})^3$
$\large\frac{d_{fcc}}{d_{bcc}} = \large\frac{4\times(3.0\times10^{-8})^3}{2\times(3.5\times10^{-8})^3}$
$\;\;\;\;\; = 1.259$
Hence answer is (C)