You have not stated the volume of your KOH solution. I therefore have to make some assumptions.

I will use pka1 (H2CO3) = 6.37

I further assume you have 1 liter of the KOH solution and that this volume will not change by the addition of the carbon dioxide. For the following, we have also to assume that ALL the solid carbon dioxide will react, i.e. no evaporation of any gas - tricky, but perhaps not impossible.

You start with a solution of KOH that has the pH = 13.2. From this we calculate pOH = 0.8 => [OH-] = 0.16 M (or 0.1584 M).

First, we suppose that the reaction between the solid carbon dioxide and the solution of KOH will result in only potassium hydrocarbonate:

CO2 + OH- = HCO3-

Ok, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate?

  You will need (per liter KOH): 0.16 mole.

After addition of 0.16 mole carbon dioxide we will face the same situation as if you had dissolved 0.16 mole of potassium hydrogencarbonate in 1 liter of water. The pH of such a solution will still be alkaline (pH > 8).

You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with pH = 6.37.

Accordingly, we have to add more carbon dioxide.

  The charge balance is:

[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO32-]

At pH = 6.37, both [OH-] and [CO32-] << [HCO3-]. Therefore, we re-write the charge balance as:

[H3O+] + [K+] = [HCO3-]

We know the values of : [H3O+] and [K+] and calculate:

[HCO3-] = 10^-6.37 + 0.16 (M).

We get [HCO3-] ≈ 0.16 (M).

We can now calculate [H2CO3]:

At pH = pka => [HCO3-] = [H2CO3] = 0.16 (M). => [HCO3-] + [H2CO3] = 0.32 (M).

Accordingly, in total we need 0.32 mole ≈ 0.3 mole of carbon dioxide to reach pH = 6.37. Per liter KOH solution this will correspond to 0.3 * 44 ≈ 13 g solid carbon dioxide.

You have not stated the volume of your $\ce{KOH}$ solution. I therefore have to make some assumptions.

I will use $\mathrm{p}K_\mathrm{a1} (\ce{H2CO3}) = 6.37$. I further assume you have $\pu{1 L}$ of the $\ce{KOH}$ solution and that this volume will not change by the addition of the carbon dioxide.

For the following, we have also to assume that all the solid carbon dioxide will react, i.e. no evaporation of any gas -- tricky, but perhaps not impossible.

You start with a solution of KOH that has the $\mathrm{pH} = 13.2$. From this we calculate $\mathrm{pOH}$:

$$\mathrm{pOH} = 0.8 \quad \to \quad [\ce{OH-}] = \pu{0.16 M}~(\text{or}~\pu{0.1584 M})$$

First, we suppose that the reaction between the solid carbon dioxide and the solution of KOH will result in only potassium hydrocarbonate:

$$\ce{CO2 + OH- -> HCO3-}$$

OK, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate? You will need (per liter $\ce{KOH}$): $\pu{0.16 mol}$.

After addition of $\pu{0.16 mol}$ carbon dioxide we will face the same situation as if you had dissolved $\pu{0.16 mol}$ of potassium hydrogencarbonate in $\pu{1 L}$ of water. The $\mathrm{pH}$ of such a solution will still be alkaline $(\mathrm{pH} > 8)$.

You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with $\mathrm{pH} = 6.37$.

Accordingly, we have to add more carbon dioxide. The charge balance is:

$$\ce{[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO3^2-]}$$

At $\mathrm{pH} = 6.37$, both $[\ce{OH-}]$ and $[\ce{CO3^2-}] << [\ce{HCO3-}]$. Therefore, we re-write the charge balance as:

$$\ce{[H3O+] + [K+] = [HCO3-]}$$

We know the values of $[\ce{H3O+}]$ and $[\ce{K+}]$, so we calculate:

$$[\ce{HCO3-}] = 10^{-6.37} + 0.16 \approx \pu{0.16 M}$$

We can now calculate $[\ce{H2CO3}]$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} \quad \to \quad [\ce{HCO3-}] = [\ce{H2CO3}] = \pu{0.16 M} \quad \to \quad [\ce{HCO3-}] + [\ce{H2CO3}] = \pu{0.32 M}$$

Accordingly, in total we need $\pu{0.32 mol} \approx \pu{0.3 mol}$ of carbon dioxide to reach $\mathrm{pH} = 6.37$. Per liter KOH solution this will correspond to $\pu{0.3 g} \times \pu{44 g mol-1} = \pu{13 g}$ solid carbon dioxide.