Step 1: Build one combined formula from N in rev per minute directly, instead of finding $\omega$ as a decimal first.
Angular speed in rad/s from $N$ revolutions per minute is $\omega = \dfrac{2\pi N}{60}$. For a ring, $I = MR^2$. So the rotational kinetic energy is
\[ KE = \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2\left(\frac{2\pi N}{60}\right)^2 = \frac{2\pi^2 M R^2 N^2}{3600} \]
Writing it this way lets us plug $N$, $M$ and $R$ in together, without rounding an intermediate decimal value of $\omega$.
Step 2: Plug in the numbers.
\[ M = 10 \ \text{kg}, \quad R = 1 \ \text{m}, \quad N = 210 \ \text{rev/min} \]
\[ KE = \frac{2\pi^2 (10)(1)^2(210)^2}{3600} = \frac{2\pi^2 \times 10 \times 44100}{3600} \]
Step 3: Simplify the numeric part before bringing in $\pi^2$.
\[ \frac{2 \times 10 \times 44100}{3600} = \frac{882000}{3600} = 245 \]
So $KE = 245\pi^2$ joules.
Step 4: Substitute $\pi^2 \approx 9.87$.
\[ KE = 245 \times 9.87 \approx 2418 \ \text{J} \]
Step 5: Average power over the 2 s it takes to stop.
All of this rotational energy is removed in $t = 2$ s, so the average power delivered against the ring is
\[ P_{avg} = \frac{KE}{t} = \frac{2418}{2} \approx 1209 \ \text{W} \]
Final Answer:
The closest option is
\[ \boxed{1210 \ \text{W}} \]