In a magnetic field of strength 0.10 T , a proton moves in a circle...

The radius of a charged particle moving in a uniform magnetic field is given by $r = \dfrac{mv}{qB}$.

Since the speed $v$ and magnetic field $B$ are the same for both particles, the radius is proportional to the mass-to-charge ratio:

$r \propto \dfrac{m}{q}$

For a proton, let mass be $m$ and charge be $e$. Then $r_{p} \propto \dfrac{m}{e}$.

For an $\alpha$-particle, mass is $4m$ and charge is $2e$. Then $r_{\alpha} \propto \dfrac{4m}{2e} = \dfrac{2m}{e}$.

Taking the ratio of the radii:

$\dfrac{r_{\alpha}}{r_{p}} = \dfrac{\dfrac{2m}{e}}{\dfrac{m}{e}} = 2$

Given $r_{p} = 1 \text{ cm}$, we get:

$r_{\alpha} = 2 \times 1 \text{ cm} = 2 \text{ cm}$