The four roots are $a \pm \sqrt{b \pm \sqrt{c}}$, so the sum of roots is 20 , so $a=5$. Next, we compute the sum of squares of roots: $$(a+\sqrt{b \pm \sqrt{c}})^{2}+(a-\sqrt{b \pm \sqrt{c}})^{2}=2 a^{2}+2 b \pm 2 \sqrt{c}$$ so the sum of squares of roots is $4 a^{2}+4 b$. However, from Vieta, it is $20^{2}-2 \cdot 108=184$, so $100+4 b=$ $184 \Longrightarrow b=21$. Finally, the product of roots is $$\left(a^{2}-(b+\sqrt{c})\right)\left(a^{2}-(b-\sqrt{c})\right)=\left(a^{2}-b\right)^{2}-c=16-c$$ so we have $c=7$.