The packing efficiency of the two-dimensional square unit cell shown in the given figure is:

Let a = 2√2r
Let us consider the square plane and L be the length of the square (given).
Area = L²
By applying Pythagoras theorem,
L² + L² = (2r)²
4r² = 2L²
√2L = 2r
L² = 2r²
Area = 2r²
Packing efficiency = (Area of two circles) / (Area of square) × 100
= (2 × πr²) / (2r²) × 100
= π × 100
= 3.14 × 100
= 314%

Let's reconsider the approach.
Let a be the side of the square.
Then the area of the square is a².
The diameter of each circle is a.
The radius of each circle is a/2.
The area of one circle is π(a/2)² = πa²/4.
The area of two circles is 2πa²/4 = πa²/2.
The packing efficiency is (πa²/2) / a² × 100 = π/2 × 100 ≈ 157.08%
This calculation is incorrect. The circles overlap outside the square.

Let's use a different approach:
Let r be the radius of each circle.
The diagonal of the square is 2r.
By Pythagorean theorem: a² + a² = (2r)²
2a² = 4r²
a² = 2r²
Area of the square = a² = 2r²
Area of one circle = πr²
Area of two circles = 2πr²
Packing efficiency = (2πr²) / (2r²) × 100 = π × 100 ≈ 314.16%

The provided solution is incorrect and the calculation has errors. Let's try a more accurate approach:
Let the side of the square be 'a'. The radius of each circle is a/2.
Area of the square = a²
Area of one circle = π(a/2)² = πa²/4
Area of two circles = 2 * πa²/4 = πa²/2
Packing efficiency = (Area of 2 circles / Area of square) * 100 = (πa²/2) / a² * 100 = π/2 * 100 ≈ 157.08%
This is still incorrect because it assumes the circles are fully contained within the square, which is not the case. It should be approximately 78.5%
Let a = 2√2r
Area of square = (2√2r)² = 8r²
Area of two circles = 2πr²
Packing efficiency = (2πr²/8r²) * 100 = π/4 * 100 ≈ 78.54%