For the given 2-D parallelogram unit cell in hexagonal close packing, the packing fraction will be:

Packing fraction 2-D :

From the figure, we can say,
$\text{Total number of atom in an unit cell,}{Z}_{eff}=1$

$\text{Packing faction}=\frac{\text{Area occupied by circles}}{\text{Area of unit cell}}$
Let us consider,
$\text{Edge length}=l$
Thus, $\text{radius}\left(r\right)=\frac{l}{2}$
From figure, it is clear that parallelogram has one complete circle.
Thus,
$\text{Area of circle}=\pi r^2=\pi {\left(\frac{l}{2}\right)}^2$
$\text{Area of parallelogram}=2\times \text{Area of}\Delta ABC$
From area of triangle, we know
$\text{Area of}\Delta ABC=\frac{1}{2}BC\times AD$

$A{D}^2=A{B}^2-B{D}^2$
$=l^2-\frac{l^2}{4}=\frac{3l^2}{4}$
$\therefore AD=\sqrt{3}\left(\frac{l}{2}\right)$
$\therefore \text{Area of}\Delta ABC=\frac{1}{2}(l\times \frac{\sqrt{3}l}{2})=\frac{\sqrt{3}{l}^2}{4}$
$\therefore \text{Area of parallelogram}=2\times \text{Area of}\Delta ABC$
$=2\frac{\sqrt{3}{l}^2}{4}=\frac{\sqrt{3}{l}^2}{2}$
$\therefore \text{Packing fraction}=\frac{\frac{\pi l^2}{4}}{\frac{\sqrt{3}{l}^2}{2}}=\frac{\pi }{2\sqrt{3}}=0.907$