Solution:

By the Cauchy-Schwarz inequality we have that
$$5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq (3 a+2 b)^{2}$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM inequality we have that
$$A \geq 2 \sqrt{\sqrt{\frac{a}{b(3 a+2)}} \cdot \sqrt{\frac{b}{a(2 b+3)}}} = \frac{2}{\sqrt[4]{(3 a+2)(2 b+3)}}$$
Finally, using again the AM-GM inequality, we get that
$$(3 a+2)(2 b+3) \leq \left(\frac{3 a+2 b+5}{2}\right)^{2} \leq 25$$
so $A \geq 2 / \sqrt{5}$ and the equality holds if and only if $a=b=1$.