1. Relevant reactions

   Given: $2\,\mathrm{Cu}^{2+}+4\,\mathrm{I}^- \rightarrow \mathrm{Cu}_2\mathrm{I}_2+\mathrm{I}_2$

   and iodine is titrated with hypo (thiosulfate): $\mathrm{I}_2+2\,\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\,\mathrm{I}^-+\mathrm{S}_4\mathrm{O}_6^{2-}$

2. Find moles of thiosulfate used

   Volume of hypo solution $=20\,\mathrm{mL}=0.020\,\mathrm{L}$

   Molarity of hypo solution $=0.02\,\mathrm{M}$

   So, moles of $\mathrm{S}_2\mathrm{O}_3^{2-}$ used: $n(\mathrm{S}_2\mathrm{O}_3^{2-})=M\times V=0.02\times 0.020=4\times 10^{-4}\,\mathrm{mol}$

3. Relate thiosulfate to iodine

   From $\mathrm{I}_2+2\,\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \cdots$

   $1\text{ mol }\mathrm{I}_2 \equiv 2\text{ mol }\mathrm{S}_2\mathrm{O}_3^{2-}$

   Therefore, moles of iodine formed: $n(\mathrm{I}_2)=\frac{4\times 10^{-4}}{2}=2\times 10^{-4}\,\mathrm{mol}$

4. Relate iodine to copper(II)

   From $2\,\mathrm{Cu}^{2+}+4\,\mathrm{I}^- \rightarrow \mathrm{Cu}_2\mathrm{I}_2+\mathrm{I}_2$

   $2\text{ mol }\mathrm{Cu}^{2+} \equiv 1\text{ mol }\mathrm{I}_2$

   Hence moles of $\mathrm{Cu}^{2+}$ in $10\,\mathrm{mL}$ sample: $n(\mathrm{Cu}^{2+})=2\times 2\times 10^{-4}=4\times 10^{-4}\,\mathrm{mol}$

5. Calculate molarity of copper sulfate solution

   Volume of copper sulfate solution $=10\,\mathrm{mL}=0.010\,\mathrm{L}$

   $M(\mathrm{Cu}^{2+})=\frac{n}{V}=\frac{4\times 10^{-4}}{0.010}=4\times 10^{-2}\,\mathrm{M}$

6. Nearest integer for the blank

   The molarity is of the form $\underline{\hspace{1cm}}\times 10^{-2}\,\mathrm{M}$ so the required integer is: $4$

Therefore, the answer is $4$.