jmc

We make $a+\sqrt{d}=\sqrt{53+20\sqrt{7}}$. Squaring both sides, we get: $$\begin{array}{r}{a}^2+2a\sqrt{d}+d=(a^2+d)+\sqrt{4a^2\cdot d}=53+20\sqrt{7}=53+\sqrt{2800}\end{array}$$We set the terms with radicals equal to each other, and ones without radicals equal. From this, we get that $a^2+d=53$ and $\sqrt{4a^2\cdot d}=\sqrt{2800}$, so $4a^2\cdot d=2800$. Solving, we get that $a=5$, and $d=28$.

Therefore, $\sqrt{53+20\sqrt{7}}=5+\sqrt{28}=5+2\sqrt{7}$. $a=5$, $b=2$, and $c=7$. $a+b+c=5+2+7=\boxed{14}$.