The mean free path (λ) of molecules in a gas is given by the formula:

$$\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}$$

where k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure.

At STP (standard temperature and pressure), the temperature is $$273\mathrm{~K}$$ and the pressure is $$1\mathrm{~atm}$$. We are given that the mean free path at STP is $$1500d$$. Let's denote the mean free path at $$373\mathrm{~K}$$ as λ':

$$\lambda' = \frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}$$

To find the ratio of the mean free path at $$373\mathrm{~K}$$ to that at STP, we can divide λ' by λ:

$$\frac{\lambda'}{\lambda} = \frac{\frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}{\frac{k(273\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}$$

The Boltzmann constant, pressure, and molecular diameter cancel out:

$$\frac{\lambda'}{\lambda} = \frac{373\mathrm{~K}}{273\mathrm{~K}}$$

Now, we can solve for λ':

$$\lambda' = \lambda \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$$

Substituting the given value of λ as $$1500d$$:

$$\lambda' = 1500d \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}$$

$$\lambda' \approx 2049d$$

Thus, the mean free path of the molecules at $$373\mathrm{~K}$$ while maintaining the standard pressure is approximately $$2049d$$.