To determine the mole fraction of urea in a solution where the molality is $$4.44 \ \mathrm{m}$$, we first need to understand the definitions and relationships involved.

Molality ($$\mathrm{m}$$) is given by:

$$ \mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} $$

Here, the molality is $$4.44 \ \mathrm{m}$$, which means there are $$4.44$$ moles of urea (solute) per kilogram of water (solvent).

The next step involves calculating the mole fraction. Mole fraction ($$X$$) of urea is given by:

$$ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} $$

Since we have $$4.44$$ moles of urea, let's determine the moles of water. The molar mass of water ($$\mathrm{H_2O}$$) is approximately $$18 \ \mathrm{g/mol}$$. Therefore, the moles of water in $$1 \ \mathrm{kg}$$ (or $$1000 \ \mathrm{g}$$) of water are:

$$ \text{Moles of water} = \frac{1000 \ \mathrm{g}}{18 \ \mathrm{g/mol}} \approx 55.56 \ \text{moles} $$

We then substitute these values into the mole fraction formula:

$$ X_{\text{urea}} = \frac{4.44}{4.44 + 55.56} $$

Simplifying the expression inside the denominator first:

$$ 4.44 + 55.56 = 60 $$

So, the mole fraction of urea becomes:

$$ X_{\text{urea}} = \frac{4.44}{60} $$

Dividing the values gives us:

$$ X_{\text{urea}} \approx 0.074 $$

The problem asks for the mole fraction in the form $$x \times 10^{-3}$$, so we convert our result to this form:

$$ X_{\text{urea}} = 74 \times 10^{-3} $$

Therefore, the value of $$x$$ is:

$$ \boxed{74} $$