1. Write the current equation

For a pure resistor connected to AC supply, $i = I_0 \sin \omega t$ where $I_0$ is the maximum (peak) current.

The rms current is $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

2. Find when current changes from maximum to rms

The current is maximum when $\sin \omega t = 1$ which occurs at $\omega t_1 = \frac{\pi}{2}$

Now we want the later time when current becomes $i = I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ So, $I_0 \sin \omega t_2 = \frac{I_0}{\sqrt{2}}$ $\sin \omega t_2 = \frac{1}{\sqrt{2}}$

After the maximum point, this occurs at $\omega t_2 = \frac{3\pi}{4}$

Thus, $\Delta (\omega t) = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$

Hence, $\Delta t = \frac{\pi/4}{\omega}$

3. Use $\omega = 2\pi f$

Given, $f = 50\,\text{Hz}$ so $\omega = 2\pi f = 100\pi\,\text{rad/s}$

Therefore, $\Delta t = \frac{\pi/4}{100\pi} = \frac{1}{400}\,\text{s}$ $\Delta t = 0.0025\,\text{s} = 2.5\,\text{ms}$

4. Conclusion

The time taken is $\boxed{2.5\,\text{ms}}$

So the correct option is A.