When the room temperature is 25^ ^ ^ C , an electric kettle used...

Total volume of water $= 4 \times 200 \text{ cc} = 800 \text{ cc}$.

Mass of water, $m = 800 \text{ g} = 0.8 \text{ kg}$.

Specific heat capacity of water, s = 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}.

At an initial room temperature of 25^{}^{{}^{\circ}} \text{C}, the heat required to boil the water (100^{}^{{}^{\circ}} \text{C}) is:

$Q_1 = ms\Delta T_1 = 0.8 \times 4200 \times (100 - 25) = 252000 \text{ J}$

The kettle takes $2 \text{ minutes}$ ($120 \text{ s}$) to boil this water. The power of the kettle is:

$P = \dfrac{Q_1}{t_1} = \dfrac{252000}{120} = 2100 \text{ W} = 2.1 \text{ kW}$

When the room temperature drops to 5^{}^{{}^{\circ}} \text{C}, the new heat required to boil the water is:

$Q_2 = ms\Delta T_2 = 0.8 \times 4200 \times (100 - 5) = 319200 \text{ J}$

The time taken to boil the water at 5^{}^{{}^{\circ}} \text{C} is:

$t_2 = \dfrac{Q_2}{P} = \dfrac{319200}{2100} = 152 \text{ s}$

The electrical energy consumed in this process (in $\text{kWh}$ or units) is:

$E = P \times t_2 = 2.1 \text{ kW} \times \dfrac{152}{3600} \text{ h} = \dfrac{319.2}{3600} \text{ kWh} \approx 0.08867 \text{ units}$

Given the \cos t of power is $\text{Re } 1.00$ ($100 \text{ paise}$) per unit, the total \cos t is:

\text{\cos t} = 0.08867 \times 100 \text{ paise} = 8.867 \text{ paise}

Rounding to the nearest integer, the \cos t is $9 \text{ paise}$.

Answer: $9 \text{ paise}$