The radius of the ring and the disc is given as r = 10 cm  = 0.10 m
Initial angular speed is given as ω₀ =10 π rad s^–1
The coefficient of kinetic friction is given as μ_k = 0.2

According to Newton’s second, the force of friction, f = ma

So we have,

μ_kmg= ma

Where,

a is the acceleration produced in the disc and the ring

m is the mass

Therefore, a = μ_kg    …..(1)

Using the first equation of motion, we can write,
v = u + at
= 0 + μ_kgt
= μ_kgt       …..(2)

The frictional force reduces the initial angular speed by applying torque in a perpendicularly outward direction.

Torque, T = –Iα

Where α is the Angular acceleration

μ_kmgr = –Iα

∴ α = -μ_kmgr / I    …..(3)

According to the first equation of rotational motion, we can write,

ω = ω₀ + αt

= ω₀ + (-μ_kmgr / I )t    …..(4)

Rolling starts when linear velocity has the velovity of rω

∴ v = r (ω₀ – μ_kmgrt / I )    …..(5)

On using equations (2) and (5), we have:

μ_kgt = r (ω₀ – μ_kmgrt / I )

= rω₀ – μ_kmgr²t / I    …..(6)

For the ring:

I = mr²

∴ μ_kgt = rω₀ – μ_kmgr²t / mr²

= rω₀ – μ_kgt

2μ_kgt = rω₀

∴ t = rω₀ / 2μ_kg

= ( 0.1 × 10 × 3.14) / (2 × 0.2 × 10 )  =  0.80s   ….. (7)

For the disc:

I = (1/2)mr²

∴ μ_kgt = rω₀ – μ_kmgr²t / (1/2)mr²

= rω₀ – 2μ_kgt

3μ_kgt = rω₀

∴ t = rω₀ / 3μ_kg

= ( 0.1 x 10 × 3.14) / (3 × 0.2 × 9.8 )  =  0.53s   …..(8)

The disc will start rolling before the ring as t_D > t_R