1. Given data

• Storage capacity of water cooler $=120$ litres
• Density of water $=1000\,\text{kg m}^{-3}$, so mass of water $m=120\,\text{kg}$
• Initial temperature of all water $=10^\circ \text{C}$
• Maximum allowable temperature of water fed into device $=30^\circ \text{C}$
• Heat generated by device (thermal load) $\dot Q_{\text{load}}=3\,\text{kW}=3000\,\text{W}$
• Cooler removes heat at constant rate $P$ watts
• Time of operation $t=3\,\text{h}=10800\,\text{s}$
• Specific heat of water $c=4.2\times 10^3\,\text{J kg}^{-1}\text{K}^{-1}$

2. Physical idea

The external device adds heat to the circulating water at a rate of $3000\,$W. The cooler simultaneously removes heat at a rate of $P\,$W.

So, the net heat gain rate of the total insulated water system is $3000-P$

Since the whole system is thermally insulated from surroundings, this net heat raises the temperature of the stored $120\,$kg of water.

For the device to operate for $3$ hours, the water supplied to it must never exceed $30^\circ\text{C}$. Since initially the entire water is at $10^\circ\text{C}$, the maximum allowed rise is $\Delta T=30-10=20^\circ\text{C}$

3. Maximum heat that can be absorbed by water

$Q_{\max}=mc\Delta T$

Substitute values: $Q_{\max}=120\times 4200\times 20$

$Q_{\max}=10{,}080{,}000\,\text{J}$

4. Net heat added in 3 hours

In time $t=10800\,$s, net heat added is $Q_{\text{net}}=(3000-P)\times 10800$

For the minimum $P$, this net heat should be exactly equal to the maximum permissible heat gain: $(3000-P)\times 10800 = 10{,}080{,}000$

5. Solve for $P$

$3000-P = \frac{10{,}080{,}000}{10800}$

$3000-P = 933.33$

$P = 3000-933.33 = 2066.67\,\text{W}$

Thus, $P_{\min}\approx 2067\,\text{W}$

6. Check options

• A: $1600$ ❌
• B: $2067$ ✅
• C: $2533$ ❌
• D: $3933$ ❌

Therefore, the correct option is B.