1. Write the reaction stoichiometry

Propyne adds two molecules of bromine to form $1,1,2,2$-tetrabromopropane:

\text{CH}_3\!-\text{C}\equiv\text{CH} + 2\text{Br}_2 \longrightarrow \text{CH}_3\!-\text{CBr}_2\!-\text{CHBr}_2

So,

• $2$ mol of $\text{Br}_2$ give $1$ mol of product.

2. Moles of bromine taken

Given mass of bromine = $1\,\text{g}$

Molar mass of bromine molecule:

$M(\text{Br}_2)=2\times 80=160\,\text{g mol}^{-1}$

Hence moles of bromine:

$n(\text{Br}_2)=\frac{1}{160}\,\text{mol}$

3. Theoretical moles of product

Since $2$ mol $\text{Br}_2$ produce $1$ mol product,

$n(\text{product})_{\text{theoretical}}=\frac{1/160}{2}=\frac{1}{320}\,\text{mol}$

4. Actual moles at 27% yield

Given yield = $27\%$

$n(\text{product})_{\text{actual}}=0.27\times \frac{1}{320}=\frac{27}{32000}\,\text{mol}$

5. Molar mass of $1,1,2,2$-tetrabromopropane

Formula: $\text{C}_3\text{H}_4\text{Br}_4$

Using $\text{C}=12$, $\text{H}=1$, $\text{Br}=80$:

$M = 3(12)+4(1)+4(80)=36+4+320=360\,\text{g mol}^{-1}$

6. Mass of product obtained

$m = n\times M = \frac{27}{32000}\times 360$

$m = 0.30375\,\text{g}$

This is approximately:

$0.3\,\text{g} = 3\times 10^{-1}\,\text{g}$

So the required integer is:

$\boxed{3}$

7. Comparison with stored answer

Stored correct answer = $3$

Our derived answer also = $3$, so they agree.