Let S= $9+9^2+9^3+\cdots+9^{2010}$

S=$\frac{9}{8}\times (9^{2010}-1)$

Clearly $(9^{2010}-1)$ is even and 9 is there so it is not divisible by 2, 3.

Using Fermat's theorem

$3^{p-1}\equiv 1(mod p)$ where p is prime

Now, 4020 =$2^2\times3\times5\times67$

From here we can eliminate all the factors of 4020

So $p-1 \neq 2,3,4,5,6,10,12,15,30......$ and so on

So $p \neq 3,4,5,6,11,13,16,31,68..........$

So, 17 is the smallest prime number left.

So $9^{17-1}\equiv 1(mod p)$

$9^{2000}\equiv 1(mod p)$

$9^{2010}-1\equiv 9^{10}-1(mod p)$

It can be solved further but the answer is $ \boxed{17} $