Between Mars' and Jupiter's orbits, the asteroid belt revolves around the sun. A single asteroid...

Given values:
${\displaystyle T_1=5years=5\cdot 365\cdot 24\cdot 3600s=1.5768\cdot {10}^{8}s}$
${\displaystyle G=6.67\cdot {10}^{-11}\frac{Nk{g}^2}{k{g}^2}}$
${\displaystyle m_{sun}=1.99\cdot {10}^{30}kg}$
The fravitational force between the sun and the orbiting asteroid is:
${\displaystyle F=G\frac{m_{sun}{m}_a}{r_1^2}}$
The asteroid is affected by centripetal acceleration because it moves in a circular orbit
${\displaystyle F=m{a}_c}$,
${\displaystyle a_c=\frac{v^2}{r}}$
According to the Newtons

Using Kepler's 3rd law which is: ${\displaystyle T^2=\frac{4{\pi }^2{r}^3}{GM}}$
Solved for r:
${\displaystyle r={\left[\frac{GM{T}^2}{4{\pi }^2}\right]}^{\frac{1}{3}}}$
Where G is the universal gravitational constantM is the mass of the sun, T is the asteroid's period in seconds, and r is the radius of the orbit.
Change 5.00 years to seconds :
5.00 years=5.00 years (365days/year)(24.0hours/day)(${\displaystyle 6\dots =1.58\times {10}^{8}s}$
The radius of the orbit then is computed:
${\displaystyle r={\left[\frac{(6.67\times {10}^{-11}N\cdot \frac{m}{k{g}^2})(1.99\times {10}^{30}kg){(1.58\times {10}^{8}s)}^2}{4{\pi }^2}\right]}^{\frac{1}{3}}=4.38\times {10}^{11}m}$