Answer
$r_i=816m$
Work Step by Step
In artificial gravity, the centripetal acceleration provides the artificial gravitational acceleration. In other words, $$g=a_c=\frac{v^2}{r}$$
The speed in 2 rings is not the same, but the period in 2 rings is the same. So, we should rewrite $v$ into $T$ for comparison: $$g=\frac{\Big(\frac{2\pi r}{T}\Big)^2}{r}=\frac{4\pi^2r^2}{rT^2}=\frac{4\pi^2r}{T^2}$$We know $g_e=9.8m/s^2$ and $g_m=3.72m/s^2$ and $r_0=2150m$
To find $r_i$, the ring that represents $g_m$, we take $$\frac{g_m}{g_e}=\frac{\frac{4\pi^2r_i}{T^2}}{\frac{4\pi^2r_o}{T^2}}=\frac{r_i}{r_o}$$ $$r_i=\frac{g_mr_o}{g_e}=816m$$
