1. Reactions involved

When $MnO_2$ reacts with hydrochloric acid:

$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$

Thus, $1$ mole of $MnO_2$ liberates $1$ mole of $Cl_2$.

The liberated chlorine reacts with potassium iodide:

$Cl_2 + 2KI \rightarrow 2KCl + I_2$

Thus, $1$ mole of $Cl_2$ liberates $1$ mole of $I_2$.

The iodine is titrated with sodium thiosulfate:

$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$

Thus, $1$ mole of $I_2$ requires $2$ moles of $Na_2S_2O_3$.

2. Moles of $Na_2S_2O_3$ used

Volume of $Na_2S_2O_3$ solution $= 60.0\text{ mL} = 0.0600\text{ L}$

Molarity $= 0.1\text{ M}$

$\text{moles of } Na_2S_2O_3 = M \times V = 0.1 \times 0.0600 = 0.0060$

3. Moles of $I_2$ liberated

From

$I_2 : Na_2S_2O_3 = 1:2$

so,

$\text{moles of } I_2 = \frac{0.0060}{2} = 0.0030$

4. Moles of $Cl_2$ produced

Since

$Cl_2 : I_2 = 1:1$

$\text{moles of } Cl_2 = 0.0030$

5. Moles of $MnO_2$ in sample

Since

$MnO_2 : Cl_2 = 1:1$

$\text{moles of } MnO_2 = 0.0030$

6. Mass of $MnO_2$

Molar mass of $MnO_2$:

$55 + 2(16) = 87\text{ g mol}^{-1}$

Therefore,

$\text{mass of } MnO_2 = 0.0030 \times 87 = 0.261\text{ g}$

7. Percentage of $MnO_2$ in 2.0 g sample

$\%MnO_2 = \frac{0.261}{2.0} \times 100 = 13.05\%$

Nearest integer:

$\boxed{13}$

8. Comparison with stored answer

Stored correct answer = $13$

Our derived answer = $13$

So, the answer agrees with the stored correct answer.