To find the ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water, we first need to understand how common ion effect influences solubility.

Let's denote the solubility product constant of AgCl as $K_{sp}$.

Given $K_{sp}$ of AgCl = $10^{-10}$.

First, we calculate the solubility of AgCl in pure water:

In water, the dissociation of AgCl can be represented as:

$$\text{AgCl (s)} \leftrightarrow \text{Ag}^{+} \text{(aq)} + \text{Cl}^{-} \text{(aq)}$$

If $s$ is the solubility of AgCl in water, then

$[Ag^{+}] = s$

$[Cl^{-}] = s$

Hence, the solubility product $K_{sp}$ can be written as:

$$K_{sp} = [Ag^{+}] [Cl^{-}]$$

$$K_{sp} = s \cdot s = s^{2}$$

So,

$s^{2} = 10^{-10}$

$s = 10^{-5}$

Now, let's consider the solubility of AgCl in 0.1 M KCl solution. Because of the common ion effect, the presence of $Cl^{-}$ ions from KCl will suppress the solubility of AgCl.

Here, $[Cl^{-}]$ from KCl is 0.1 M. Let the new solubility of AgCl in this solution be $s'$.

Then,

$[Ag^{+}] = s'$

$$[Cl^{-}] = 0.1 + s' \approx 0.1$$

Since $s'$ is much smaller than 0.1 M, we can approximate:

$$K_{sp} = [Ag^{+}] [Cl^{-}]$$

$$10^{-10} = s' \times 0.1$$

$$s' = \frac{10^{-10}}{0.1}$$

$s' = 10^{-9}$

Finally, we find the ratio of solubility in 0.1 M KCl to that in pure water:

$$\text{Ratio} = \frac{s'}{s} = \frac{10^{-9}}{10^{-5}} = 10^{-4}$$

Therefore, the correct answer is:

Option A

10$^{-4}$