When two conductors are connected by a conducting wire, they reach the same potential, so $V_1 = V_2$. Using $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$, we have $\frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1} = \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R_2}$. Since surface charge density $\sigma = \frac{Q}{4\pi R^2}$, we substitute $Q = \sigma 4\pi R^2$ to get $\frac{\sigma_1 4\pi R_1^2}{R_1} = \frac{\sigma_2 4\pi R_2^2}{R_2}$, which simplifies to $\sigma_1 R_1 = \sigma_2 R_2$, hence $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$.