The elevation in boiling point for 1 molal solution of non-volatile solute A is $$3 \text{ K}$$. The depression in freezing point for 2 molal solution of A in the same solvent is $$6 \text{ K}$$. The ratio of $$K_b$$ and $$K_f$$ i.e., $$K_b/K_f$$ is $$1:X$$. The value of $$X$$ is ______.

We are given that the elevation in boiling point for a 1 molal solution of non-volatile solute A is $$3\text{ K}$$ and the depression in freezing point for a 2 molal solution of A in the same solvent is $$6\text{ K}$$.

Since the elevation in boiling point is related by $$\Delta T_b = K_b \cdot m$$, substituting $$m = 1$$ molal gives $$3 = K_b \times 1$$, which yields $$K_b = 3\text{ K kg/mol}$$.

Similarly, because the depression in freezing point follows $$\Delta T_f = K_f \cdot m$$, substituting $$m = 2$$ molal leads to $$6 = K_f \times 2$$, and hence $$K_f = 3\text{ K kg/mol}$$.

Therefore, the ratio is $$\frac{K_b}{K_f} = \frac{3}{3} = 1$$.

Thus, $$K_b / K_f = 1 : 1$$, which means $$X = 1$$.

Therefore, the answer is $$\boxed{1}$$.