Other answers here use symmetry. This is very useful, but if the symmetry of the problem is not clear to you, this can always be calculated with conditional probability. Let's denote the event $A$ as "the first ball is #3", and $B$ "the second ball is #3". So conditional probability tells you that $\begin{eqnarray} P(B)&=&P(B|A)\cdot P(A)+P(B|A^c)\cdot P(A^c)\\ &=&P(B|A)\cdot P(A)+P(B|A^c)\cdot(1- P(A)) \end{eqnarray} $ ($A^c$ is the complement of $A$, i.e. the event "not A"). Now, the probability of $A$ is, as you probably know, $P(A)=\frac{1}{10}$. But given $A$, the probability of $B$ is $0$, because #3 was already taken out. The probability of $B$ given $A^c$ is, as you correctly stated, $P(B|A^c)=\frac{1}{9}$. So this gives
$P(B)=0\cdot \frac{1}{10} + \frac{1}{9}\cdot\left(1-\frac{1}{10}\right)=\frac{1}{10}$