Two iron solid discs of negligible thickness have radii $$R_{1}$$ and $$R_{2}$$ and moment of intertia $$I_{1}$$ and $$I_{2}$$, respectively. For $$R_{2}=2R_{1}$$, the ratio of $$I_{1}$$ and $$I_{2}$$ would be $$1/x$$, where x = __________.

Two iron solid discs have radii $$R_1$$ and $$R_2$$ with $$R_2 = 2R_1$$. We first determine the ratio $$I_1/I_2$$.

Next, we recall that the moment of inertia of a solid disc is given by
$$I = \frac{1}{2}MR^2$$

Since both discs are made of iron (same density $$\rho$$) and have negligible thickness $$t$$, their mass can be expressed as
$$M = \rho \cdot \pi R^2 \cdot t$$ so $$M \propto R^2$$.

Substituting into the inertia formula gives
$$I = \frac{1}{2}MR^2 = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{\rho \pi t}{2} R^4$$ and hence $$I \propto R^4$$.

It follows that
$$\frac{I_1}{I_2} = \frac{R_1^4}{R_2^4} = \frac{R_1^4}{(2R_1)^4} = \frac{R_1^4}{16R_1^4} = \frac{1}{16}\,. $$

Therefore, $$I_1/I_2 = 1/x$$ where $$x = \boxed{16}$$.