jmc

Since 2 is not a multiple of 3 or 7, $f(2)=2+3=5$ and we want to find an $a$ where $f^a(2)=5$. So, we keep track of how many times we evaluate $f$ of our previous result until we get 5. $$\begin{array}{rl}f(2)& =5\\ f(f(2))& =f(5)=5+3=85\text{ is not a multiple of 3 or 7.}\\ f(f(f(2)))& =f(8)=8+3=118\text{ is not a multiple of 3 or 7.}\\ f^4(2)& =11+3=1411\text{ is not a multiple of 3 or 7.}\\ f^5(2)& =3\cdot 14=4214\text{ is a multiple of 7.}\\ f^6(2)& =\frac{42}{21}=242\text{ is a multiple of 3 and 7.}\\ f^7(2)& =2+3=52\text{ is not a multiple of 3 or 7.}\end{array}$$So the least $a>1$ for which $f^a(2)=f(2)$ is $a=\boxed{7}$.