If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2:1 and their enthalpy of vaporizations are in the ratio 1:2, then the boiling point elevation constant of X is m times the boiling point elevation constant of Y. The value of m is (nearest integer).

The boiling point elevation constant (ebullioscopic constant) is given by:

$$K_b = \frac{R T_b^2 M}{1000 \, \Delta H_{\text{vap}}}$$

where $$R$$ is the gas constant, $$T_b$$ is the boiling point (in Kelvin), $$M$$ is the molar mass of the solvent, and $$\Delta H_{\text{vap}}$$ is the enthalpy of vaporization.

Since both solvents X and Y have the same molecular weight, $$M$$ and $$R$$ cancel in the ratio:

$$\frac{K_{b,X}}{K_{b,Y}} = \frac{T_{b,X}^2}{T_{b,Y}^2} \times \frac{\Delta H_{\text{vap},Y}}{\Delta H_{\text{vap},X}}$$

We have $$\frac{T_{b,X}}{T_{b,Y}} = \frac{2}{1}$$ and $$\frac{\Delta H_{\text{vap},X}}{\Delta H_{\text{vap},Y}} = \frac{1}{2}$$. Substituting:

$$\frac{K_{b,X}}{K_{b,Y}} = \left(\frac{2}{1}\right)^2 \times \frac{2}{1} = 4 \times 2 = 8$$

So $$K_{b,X} = 8 \times K_{b,Y}$$, meaning $$m = 8$$.

So, the answer is $$8$$.