When an electric kettle is used to heat water, the time taken to boil the water depends on the power of the heating element. The power supplied to the heating element is inversely proportional to the heating time. If we want to reduce the boiling time, the power needs to be increased. The power of the heating element is given by:

$$P = \frac{V^2}{R}$$

where $$P$$ is the power, $$V$$ is the voltage, and $$R$$ is the resistance of the heating element. The resistance $$R$$ of the heating element is proportional to its length $$L$$ while the material and cross-sectional area remain constant.

So, we can write:

$$R \propto L$$

To achieve boiling in 15 minutes instead of 20 minutes, the power needs to increase, which implies the resistance must decrease. Let the initial length of the heating element be $$L_0$$, and let the new length needed be $$L$$. The time taken to heat is inversely proportional to the power:

$$\frac{T_1}{T_2} = \frac{P_2}{P_1}$$

Given that:

$$T_1 = 20 \text{ minutes}$$

$$T_2 = 15 \text{ minutes}$$

We need to find the ratio:

$$\frac{20}{15} = \frac{P_2}{P_1}$$

Simplifying:

$$\frac{4}{3} = \frac{P_2}{P_1}$$

The power is inversely proportional to the resistance:

$$\frac{P_2}{P_1} = \frac{R_1}{R_2}$$

Thus:

$$\frac{4}{3} = \frac{R_1}{R_2}$$

Since resistance is proportional to length:

$$\frac{R_1}{R_2} = \frac{L_1}{L_2}$$

Hence:

$$\frac{4}{3} = \frac{L_1}{L_2}$$

Simplifying, we find:

$$L_2 = \frac{3}{4} L_1$$

This means the length of the heating element should be decreased to $$\frac{3}{4}$$ of its initial length.

Thus, the correct option is:

Option C: decreased, $$\frac{3}{4}$$