The first and second ionization constants of a weak dibasic acid $H_2 A$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. $0.1$ mol of $H_2 A$ was dissolved in $1$ L of $0.1$ M HCl solution. The concentration of $HA^-$ in the resultant solution is:

Given:
Weak dibasic acid $H_2A$ with $K_{a1} = 8.1 \times 10^{-8}$ and $K_{a2} = 1.0 \times 10^{-13}$
Initial concentration: $[H_2A]_0 = 0.1\text{ M}$
The solution also contains $0.1\text{ M HCl}$, which dissociates completely to give $[H^+]_0 = 0.1\text{ M}$.

The main source of $HA^-$ is the first dissociation:
$H_2A \rightleftharpoons H^+ + HA^-$

Let $x$ be the amount of $H_2A$ that dissociates. Then at equilibrium:
$[H_2A] = 0.1 - x$
$[H^+] = 0.1 + x$
$[HA^-] = x$

Applying the equilibrium expression:
$K_{a1} = \dfrac{[H^+][HA^-]}{[H_2A]}$

$8.1 \times 10^{-8} = \dfrac{(0.1 + x)(x)}{0.1 - x}$

Since $K_{a1}$ is very small and the common $H^+$ ion from HCl suppresses dissociation, $x \ll 0.1$. Hence:
$0.1 - x \approx 0.1$ and $0.1 + x \approx 0.1$

$8.1 \times 10^{-8} \approx \dfrac{(0.1)(x)}{0.1} = x$

Therefore:
$[HA^-] = x = 8.1 \times 10^{-8}\text{ M}$

The second dissociation ($K_{a2} = 1.0 \times 10^{-13}$) is negligible compared to the first, so the $HA^-$ consumed in that step is insignificant.

Hence, the correct option is $(3)\ 8.1 \times 10^{-8}\text{ M}$.