Why do you think the entropy of 0 is incorrect? It intuitively makes sense, as there is no uncertainty about the base at position 3, and thus there is no entropy.
However, what is plotted in a sequence logo isn't the entropy, but rather a measure of the "decrease in uncertainty" as the sequence is aligned. This is calculated by taking the entropy at this position if we randomly aligned sequences ($H_g(L)$), and subtracting from it the entropy of the alignment ($H_g(s)$):
$$
R_{sequence}=H_g(L) - H_s(L)
$$
The entropy at position 3 based on a random alignment is calculated by assuming there are 4 equally likely events (one for each base), and thus:
$$
\begin{align*}
H_g(L) & = -((1/4 \times -2) + (1/4 \times -2) + (1/4 \times -2) + (1/4 \times -2)) \\
H_g(L) & = 2
\end{align*}
$$
Notably, $H_g(L)$ will always be 2 when dealing with nucleotide sequences.
So in your example, we have:
$$
\begin{align*}
R_{sequence}&=H_g(L) - H_s(L) \\
R_{sequence}&=2 - 0 \\
R_{sequence}&=2 \text{ bits of entropy}
\end{align*}
$$
Finally, to work out the height of each base at each position in the logo, we multiply the frequency of that base by the overall information gain $R_{sequence}$. In this case the frequency of the base A at position 3 is of course 1:
$$
\begin {align*}
H(b,l) &=f(b, l) \times R_{sequence} \\
H(A, 3) &= f(a, 3) \times 2 \\
&= 1 \times 2 \\
&= 2
\end{align*}
$$
Source: Sequence logos: a new way to display consensus sequences.
(Thomas D. Schneider, R. Michael Stephens, 1990)
Note: I've excluded the correction factor $e(n)$ from the overall information-gain calculation for the sake of simplicity. Refer to the paper above for an explanation of this.