You start with three circles initially, that are not going to be counted initially.
Then you add the inner and outer tangent circles : $a_0=2$.
From then on every circle from stage $n$ is going to be surrounded by three new circles at stage $n+1$, therefore the amount of circles added at stage $n+1$ is $$a_{n+1}=3a_n$$
Given that you start $a_0=2$ it follows that $a_n=2\cdot 3\cdot\ldots\cdot 3=2\cdot3^{n}$
Now the **total** number of circles at stage $n$ is $$a_0+\ldots+a_n=2\cdot(3^0+\ldots+3^n)$$ which is $$2\cdot\frac{3^{n+1}-1}{3-1}$$that is $3^{n+1}-1$.
Finally, add in the three initial circles: $3^{n+1}-1+3=3^{n+1}+2$.