1. Given data

• Water in calorimeter: $180\,\text{g}$ at $25^\circ\text{C}$
• Water equivalent of calorimeter: $20\,\text{g}$
• Steam added: $m\,\text{g}$ at $100^\circ\text{C}$
• Final temperature: $31^\circ\text{C}$
• Latent heat of steam: $L = 540\,\text{cal g}^{-1}$
• Specific heat of water: $c = 1\,\text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$

2. Heat gained by water + calorimeter

Since calorimeter water equivalent is $20\,\text{g}$, total effective mass heated is $180 + 20 = 200\,\text{g}.$

Temperature rise: $31 - 25 = 6^\circ\text{C}.$

So heat gained is $Q_\text{gained} = 200 \times 1 \times 6 = 1200\,\text{cal}.$

3. Heat released by steam

Each gram of steam at $100^\circ\text{C}$ first condenses to water at $100^\circ\text{C}$, releasing $540\,\text{cal/g}.$

Then this water cools from $100^\circ\text{C}$ to $31^\circ\text{C}$, releasing $1 \times (100-31)=69\,\text{cal/g}.$

So total heat released per gram of steam is $540 + 69 = 609\,\text{cal/g}.$

Thus total heat released by $m$ grams steam is $Q_\text{released} = 609m.$

4. Apply principle of calorimetry

$Q_\text{released} = Q_\text{gained}$

$609m = 1200$

$m = \frac{1200}{609} \approx 1.97\,\text{g}.$

5. Closest option

$m \approx 2\,\text{g}.$

So the correct option is B.