jmc

Expanding $(\sqrt{a}-b{)}^3,$ we have $$\begin{array}{rl}(\sqrt{a}-b{)}^3& =a\sqrt{a}-3ab+3b^2\sqrt{a}-b^3\\ & =(a+3b^2)\sqrt{a}+(-3ab-b^3).\end{array}$$Since $a$ and $b$ are integers, we must have $$\begin{array}{rl}(a+3b^2)\sqrt{a}& =\sqrt{2700},\\ -3ab-b^3& =-37.\end{array}$$The second equation factors as $b(3a+b^2)=37.$ Since $37$ is a prime, we must have $b=37$ or $b=1.$ If $b=37,$ then $3a+b^2=1,$ which has no positive integer solutions for $a.$ Therefore, $b=1,$ and we have $3a+b^2=37,$ which gives $a=12.$

Indeed, $(a,b)=(12,1)$ also satisfies the first equation: $$(a+3b^2)\sqrt{a}=(12+3\cdot 1^2)\sqrt{12}=15\sqrt{12}=\sqrt{2700}.$$Therefore, $a+b=12+1=\boxed{13}.$