Answer
Approximately 30 drops of that 12 M HCl solution.
Work Step by Step
1. Draw the ICE table for the equilibrium of only 0.100 M CH3COOH:
$$\begin{vmatrix}
Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^+ ]\\
Initial& 0.100 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& x& x\\
\end{vmatrix}$$
2. Now, we have to find an initial concentration of hydronium ion, given by the HCl, that makes $x = [CH_3COO^-] = 1.0 \times 10^{-4}$. Let's call that "y".
$$\begin{vmatrix}
Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^+ ]\\
Initial& 0.100 & 0 & y \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& x& y + x\\
\end{vmatrix}$$
$$K_a = 1.8 \times 10^{-5} = \frac{x(y+x)}{0.100 - x}$$
We know that x = $1.0 \times 10^{-4}$
$$ 1.8 \times 10^{-5} = \frac{(1.0 \times 10^{-4})(y+1.0 \times 10^{-4})}{0.100 - 1.0 \times 10^{-4}} = \frac{(1.0 \times 10^{-4})(y + 1.0 \times 10^{-4})}{0.100}$$
3. Solve for y:
y = 0.0179 M
Therefore, we need to start with $0.0179 M$ of $H_3O^+$ in order to get this concentration at equilibrium, which means 0.0179 M of HCl.
$$1.00 \space L \times \frac{0.0179 \space mol \space HCl}{1 \space L} = 0.0179 \space mol \space HCl$$
$$0.0179 \space mol \space HCl \times \frac{1 \space L}{12 \space mol \space HCl} = 1.49 \times 10^{-3} \space L = 1.49 mL \approx 1.5 mL $$
$$1.5 \space mL \times \frac{1 \space drop}{0.050 \space mL} = 30 \space drops$$
