Stacked semicircles in a quarter circle.

Using the above diagram $1 = 5R{1}^2 \implies R_{1} = \dfrac{1}{\sqrt{5}}$ and $R{1} + R_{2} = \dfrac{\sqrt{5}R_{2} + 1}{\sqrt{5}} \implies$

$\dfrac{(\sqrt{5}R_{2} + 1)^2}{5} + 4R_{2}^2 = 1 \implies 25R_{2}^2 + 2\sqrt{5}R_{2} - 4 = 0 \implies R_{2} = \dfrac{\sqrt{105} - \sqrt{5}}{25}$ dropping the negative root

and $(R_{1} + R_{2} + R_{3})^2 + 4R_{3}^2 = 1 \implies 5R_{3}^2 + 2(R_{1} + R_{2})R_{3} + (R_{1} + R_{2})^2 - 1 = 0 \implies$

$R_{3} = \dfrac{-(R_{1} + R_{2}) + \sqrt{5 - 4(R_{1} + R_{2})^2}}{5}$ , where $R_{1} + R_{2} = \dfrac{\sqrt{21} + 4}{5\sqrt{5}} \implies$

$(R_{1} + R_{2})^2 = \dfrac{37 + 8\sqrt{21}}{125} \implies 5 - 4(R_{1} + R_{2})^2 = \dfrac{477 - 32\sqrt{21}}{125} \implies$

$R_{3} = \dfrac{\sqrt{477 - 32\sqrt{21}} - \sqrt{21} - 4}{25\sqrt{5}} = \dfrac{\sqrt{477 - 2^5\sqrt{21}} - \sqrt{21} - 2^2}{5^2\sqrt{5}} =$

$\dfrac{\sqrt{a - (b^c)\sqrt{d}} - \sqrt{d} - b^b}{(c^b)\sqrt{c}} \implies a + b + c + d = \boxed{505}$ .