After being accelerated through a potential difference of 500 V ,...

The kinetic energy acquired by an ion accelerated through a potential difference $V$ is $K = qV$.

The radius of the circular path described by an ion in a uniform magnetic field is given by $R = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB} = \dfrac{1}{B} \sqrt{\dfrac{2mV}{q}}$.

Given values are:
$V = 500 \text{ V}$
$B = 20.0 \times 10^{-3} \text{ T}$
$q = 1.6 \times 10^{-19} \text{ C}$
$m = A \times 1.6 \times 10^{-27} \text{ kg}$

Substituting these values into the formula for radius:
$R = \dfrac{1}{20.0 \times 10^{-3}} \sqrt{\dfrac{2 \times A \times 1.6 \times 10^{-27} \times 500}{1.6 \times 10^{-19}}}$

$R = 50 \sqrt{A \times 10^{-5}} \text{ m}$

$R = 0.05 \sqrt{10A} \text{ m} = 5 \sqrt{10A} \text{ cm}$

For the isotope with mass number $A = 57$:
$R_{57} = 5 \sqrt{570} \approx 5 \times 23.87 \text{ cm} \approx 119.35 \text{ cm} \approx 119 \text{ cm}$

For the isotope with mass number $A = 58$:
$R_{58} = 5 \sqrt{580} \approx 5 \times 24.08 \text{ cm} \approx 120.4 \text{ cm} \approx 120 \text{ cm}$

The radii of the circular paths are approximately $119 \text{ cm}$ and $120 \text{ cm}$.