Answer
(a) $r_A=912m$
(b) $r_B=228m$
(c) $g_B=2.5m/s^2$
Work Step by Step
(a) To create artificial gravity, the centripetal force $F_c$ acts as the gravitational force pulling objects to the center. In other words, $$F_c=mg$$ $$\frac{mv^2}{r}=mg$$ $$g=\frac{v^2}{r}$$
Here we know period $T$ but not rotation speed $v$. But we remember that $v=\frac{2\pi r}{T}$, so $$g=\frac{\frac{4\pi^2r^2}{T^2}}{r}=\frac{4\pi^2r}{T^2}$$ $$r=\frac{gT^2}{4\pi^2}$$
The space station is rotating at 1.00 rpm, which means its period $T=1min=60s$. We also know $g_A=10m/s^2$. From here, we can find $r_A$
$$r_A=912m$$
(b) Since $r_A/r_B=4$, we have $r_B=228m$
(c) To find $g_B$, we rely on the equation deduced in (a): $g=\frac{4\pi^2r}{T^2}$
The period $T$ for chamber A and B is the same. So, $$\frac{g_B}{g_A}=\frac{\frac{4\pi^2r_B}{T^2}}{\frac{4\pi^2r_A}{T^2}}=\frac{r_B}{r_A}=\frac{1}{4}$$ $$g_B=\frac{1}{4}g_A=2.5m/s^2$$
