Let $I$ be the moment of inertia, $\rho$ be the density, $R$ be the radius, and $t$ be the thickness of the discs.

The moment of inertia of a disc about its axis is given by $I = \frac{1}{2}MR^2$, where $M$ is the mass of the disc.

The mass of the disc can be expressed as $M = \rho V = \rho \pi R^2 t$, where $V$ is the volume of the disc.

Substituting the expression for $M$ into the moment of inertia formula, we get $I = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{1}{2}\pi \rho R^4 t$.

Since the moment of inertia is the same for both discs, we have $I_1 = I_2$.

Therefore, $\frac{1}{2}\pi \rho R_1^4 t_1 = \frac{1}{2}\pi \rho R_2^4 t_2$.

Since the density is the same for both discs, we can cancel out $\frac{1}{2}\pi \rho$ from both sides, giving us $R_1^4 t_1 = R_2^4 t_2$.

We are given that $\frac{R_1}{R_2} = \frac{1}{2}$. Therefore, $R_1 = \frac{1}{2}R_2$.

Substituting this into the equation $R_1^4 t_1 = R_2^4 t_2$, we get $(\frac{1}{2}R_2)^4 t_1 = R_2^4 t_2$.

Simplifying, we have $\frac{1}{16}R_2^4 t_1 = R_2^4 t_2$.

Dividing both sides by $R_2^4$, we get $\frac{1}{16}t_1 = t_2$.

Therefore, $\frac{t_1}{t_2} = 16$.