1. Given reaction

$5\mathrm{I}^- + \mathrm{IO}_3^- + 6\mathrm{H}^+ \rightarrow 3\mathrm{I}_2 + 3\mathrm{H}_2\mathrm{O}$

We evaluate each statement using the stoichiometric coefficients.

2. Moles of KI in the given solution volumes

The KI solution is $0.1\,\mathrm{M}$.

• For $200\,\mathrm{mL} = 0.2\,\mathrm{L}$: $n(\mathrm{KI}) = M\times V = 0.1\times 0.2 = 0.02\,\mathrm{mol}$ Since KI gives one $\mathrm{I}^-$ per mole, $n(\mathrm{I}^-) = 0.02\,\mathrm{mol}$

• For $0.5\,\mathrm{L}$: $n(\mathrm{KI}) = 0.1\times 0.5 = 0.05\,\mathrm{mol}$ Hence, $n(\mathrm{I}^-) = 0.05\,\mathrm{mol}$

3. Check statement (A)

From the reaction, $5\mathrm{I}^- : 1\mathrm{IO}_3^-$

So for $0.02$ mol $\mathrm{I}^-$ required moles of $\mathrm{IO}_3^-$ are $n(\mathrm{IO}_3^-) = \frac{0.02}{5} = 0.004\,\mathrm{mol}$

Since $\mathrm{KIO}_3$ provides one $\mathrm{IO}_3^-$ per mole, $n(\mathrm{KIO}_3)=0.004\,\mathrm{mol}$

So (A) is correct.

4. Check statement (B)

From the reaction, $5\mathrm{I}^- : 6\mathrm{H}^+$

For $0.02$ mol $\mathrm{I}^-$, $n(\mathrm{H}^+) = \frac{6}{5}\times 0.02 = 0.024\,\mathrm{mol}$

Now $\mathrm{H}_2\mathrm{SO}_4$ gives $2\mathrm{H}^+$ per mole, so required moles of sulfuric acid are $n(\mathrm{H}_2\mathrm{SO}_4)=\frac{0.024}{2}=0.012\,\mathrm{mol}$

But the statement says $0.006$ mol.

So (B) is incorrect.

5. Check statement (C)

From the reaction, $5\mathrm{I}^- \rightarrow 3\mathrm{I}_2$

For $0.05$ mol $\mathrm{I}^-$, $n(\mathrm{I}_2)=\frac{3}{5}\times 0.05=0.03\,\mathrm{mol}$

But the statement says $0.005$ mol.

So (C) is incorrect.

6. Check statement (D)

In $\mathrm{KIO}_3$, iodine in $\mathrm{IO}_3^-$ changes from oxidation state $+5$ to $0$ in $\mathrm{I}_2$.

Thus, 1 mole of $\mathrm{KIO}_3$ gains $5$ electrons.

Therefore, n-factor of $\mathrm{KIO}_3$ is $5$, so equivalent weight is $\text{Equivalent weight} = \frac{\text{Molecular weight}}{5}$

So (D) is correct.

7. Final conclusion

Correct statements are: $\boxed{(A) \text{ and } (D)}$

Hence the correct option is $\boxed{\text{D}}$