Last digit). Then $A=$ $\qquad$

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2. $\frac{2}{99}$.

From $3^{2} \equiv 9(\bmod 10), 3^{4} \equiv 1(\bmod 10), 3^{6} \equiv 9(\bmod 10), 3^{8} \equiv 1(\bmod 10)$, we get $a_{1}=0$, $a_{2}=2, a_{3}=0, a_{4}=2, \cdots$ In general, we have $a_{2 k-1}=0, a_{2 k}=2\left(k \in \mathbf{N}^{*}\right)$. Therefore, $A=0 . \dot{0} \dot{2}=\frac{2}{99}$.