1. Relate heating time to power

For the same quantity of water, the heat required to boil it is fixed:

$Q = \text{constant}$

Since electrical heating gives

$Q = Pt$

we get

$t \propto \frac{1}{P}$

So, if the time changes from $20$ min to $15$ min,

$\frac{P_2}{P_1} = \frac{t_1}{t_2} = \frac{20}{15} = \frac{4}{3}$

Thus, the new power must be $\frac{4}{3}$ times the original power.

2. Use the relation between power and resistance

The kettle is connected to the same mains supply, so voltage is constant.

For constant voltage,

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

Hence,

$P \propto \frac{1}{R}$

Therefore,

$\frac{R_2}{R_1} = \frac{P_1}{P_2} = \frac{1}{4/3} = \frac{3}{4}$

So the new resistance should be $\frac{3}{4}$ of the original resistance.

3. Relate resistance to length of heating element

For the same material and same cross-sectional area,

$R = \rho \frac{L}{A}$

Thus,

$R \propto L$

So,

$\frac{L_2}{L_1} = \frac{R_2}{R_1} = \frac{3}{4}$

This means the length must be reduced to $\frac{3}{4}$ of its initial length.

4. Interpret the option

The length should be decreased to $\frac{3}{4}$ times its initial length.

So the correct option is:

$\boxed{\text{C: decreased, } \frac{3}{4}}$