Given, m=1 kg
$\begin{aligned}
& \omega_{\mathrm{i}}=1800 \mathrm{rpm}=1800 \times \frac{2 \pi}{60}=60 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \
& \omega_{\mathrm{f}}=2100 \mathrm{rpm}=2100 \times \frac{2 \pi}{60}=70 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \
& \tau_{e x t}=25 \pi \mathrm{Nm} \
& \mathrm{t}=40 \mathrm{sec}
\end{aligned}$
Using equation of motion
$\begin{aligned}
& \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t} \
& 70 \pi=60 \pi+\alpha(40) \
& \alpha=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}^2
\end{aligned}$
Also, τ=Iα
τ=4mR2α
$\begin{aligned}
& 25 \pi=\frac{1 \times \mathrm{R}^2}{4} \times \frac{\pi}{4} \
& \mathrm{R}=20 \mathrm{~m}
\end{aligned}$
Hence, diameter of disk =2R=2×20=40 m