1. Magnetic field at the center due to the large loop

For a circular loop of radius $b$ carrying current $I$, the magnetic field at its center is

$B = \frac{\mu_0 I}{2b}$

Here,

$I = I_0 \cos(\omega t)$

So the magnetic field at the location of the small loop is

$B(t) = \frac{\mu_0 I_0}{2b}\cos(\omega t)$

2. Flux through the small loop

Since the small loop of radius $a$ is at the center of the large loop and $a \ll b$, the magnetic field over the small loop may be taken as nearly uniform and equal to the field at the center.

Area of the small loop:

$A = \pi a^2$

Hence magnetic flux through the small loop is

$\Phi = BA = \frac{\mu_0 I_0}{2b}\cos(\omega t) \cdot \pi a^2$

$\Phi = \frac{\pi \mu_0 I_0 a^2}{2b} \cos(\omega t)$

3. Induced emf

By Faraday's law,

$\mathcal{E} = -\frac{d\Phi}{dt}$

Differentiate:

$\mathcal{E} = -\frac{d}{dt}\left(\frac{\pi \mu_0 I_0 a^2}{2b} \cos(\omega t)\right)$

$\mathcal{E} = -\frac{\pi \mu_0 I_0 a^2}{2b}(-\omega \sin(\omega t))$

$\mathcal{E} = \frac{\pi \mu_0 I_0 a^2}{2b}\,\omega \sin(\omega t)$

Thus,

$\boxed{\mathcal{E} = \frac{\pi \mu_0 I_0}{2}\frac{a^2}{b}\,\omega \sin(\omega t)}$

4. Match with options

This matches Option A.

5. Comparison with stored answer

Stored correct answer: A
Derived answer: A

So they agree.