(2r)² = h² + x²
h² = 2r²
height = $\sqrt{2}\mathrm{r}$, 

area = 4r²

Volume = area × height
= 4r² × $\sqrt{2}\mathrm{r}\Rightarrow 4\sqrt{2}{\mathrm{r}}^3$ , 

Z = 1, $\{\frac{1}{8}\mathrm{th}\mathrm{is}\mathrm{shared}\mathrm{in}{1}^{\mathrm{st}}\mathrm{layer}\&\frac{1}{2}\mathrm{is}\mathrm{shared}\mathrm{in}{2}^{\mathrm{nd}}\mathrm{layer}\therefore \mathrm{Z}=(\frac{1}{8}\times 4+0.5)\}$ 
P.F. = $\frac{1\times 4/3{\mathrm{\pi r}}^3}{4\sqrt{2}{\mathrm{r}}^3}=\frac{\mathrm{\pi }}{3\sqrt{2}}$ = 0.75 ⇒ 75 % 

OR
As this packing also give rise to FCC lattice \ it can be directly commented that P.F. = 0.75