A hypothetical 2D solid forms a square lattice. Atoms of type X...

Let the radius of atom X be $r$. Then the radius of atom Y is $2r$.

The atoms touch along the diagonal of the square unit cell. The length of the diagonal is equal to the sum of the diameters of the atoms along it:
$\sqrt{2}a = 2r_X + 2r_Y = 2r + 2(2r) = 6r$
$\Rightarrow a = \dfrac{6r}{\sqrt{2}} = 3\sqrt{2}r$

The total area of the square unit cell is:
$A_{cell} = a^2 = (3\sqrt{2}r)^2 = 18r^2$

The effective number of atoms per 2D unit cell is:
For X (corners): $4 \times \dfrac{1}{4} = 1$
For Y (center): $1$

The total area occupied by the atoms is:
$A_{occupied} = \pi r_X^2 + \pi r_Y^2 = \pi r^2 + \pi (2r)^2 = \pi r^2 + 4\pi r^2 = 5\pi r^2$

The 2D packing fraction is:
$\text{Packing Fraction} = \dfrac{A_{occupied}}{A_{cell}} = \dfrac{5\pi r^2}{18r^2} = \dfrac{5\pi}{18}$

Answer: $\dfrac{5\pi}{18}$