1. Meter bridge balance condition

In a meter bridge, if the null point is at length $l$ from one end, then

$\frac{R_1}{R_2} = \frac{l}{100-l}$

where $R_1$ and $R_2$ are the resistances in the two gaps corresponding to the segments $l$ and $100-l$.

2. First balance: $X$ against $Y$

Given null point at $20\,\text{cm}$ from the same end.

So,

$\frac{X}{Y} = \frac{20}{80} = \frac{1}{4}$

Thus,

$Y = 4X$

This is consistent with the given condition $X<Y$.

3. Second balance: $4X$ against $Y$

Now we balance $4X$ against $Y$.

But since $Y=4X$, we get

$4X = Y$

So the two gap resistances are equal.

In a meter bridge, when resistances are equal, the balance point is at the midpoint:

$l = 50\,\text{cm}$

4. Check options

The correct option is:

$\boxed{50\,\text{cm}}$

So, Option C is correct.