Option A: $2\sqrt{10}$ units
Distance $= \sqrt{(-2\sqrt{5}-0)^2 + (0-2\sqrt{5})^2} = \sqrt{20+20} = \sqrt{40} = 2\sqrt{10}$ units.
Source: Distance Formula, Chapter 7
Apply the distance formula $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ with $(x_1,y_1)=(0,2\sqrt{5})$ and $(x_2,y_2)=(-2\sqrt{5},0)$. Each squared term gives $(\sqrt{5})^2 \times 4 = 20$, so $d=\sqrt{40}=2\sqrt{10}$. Note $2\sqrt{20}=4\sqrt{5}\neq 2\sqrt{10}$, so option C is a common distractor — simplify $\sqrt{40}$ correctly.