A long solenoid of radius 2.5 cm has 650 turns in each metre…

Step 1. The first coil. $M = \mu_0 nN_2\pi r_2^2 = 4\pi \times 10^{-7}\ \text{T m/A} \times 650\ \text{m}^{-1} \times 90 \times \pi \times (0.018\ \text{m})^2 = 7.48 \times 10^{-5}\ \text{H}$. That is $74.8$ µH.

Step 2. The emf. $|\varepsilon_2| = M\left|\dfrac{dI_1}{dt}\right| = 7.48 \times 10^{-5}\ \text{H} \times 30\ \text{A/s} = 2.24 \times 10^{-3}\ \text{V}$, that is $2.24$ mV.

Step 3. The wider coil. The field is zero outside the solenoid. So the flux through the coil uses the area of the solenoid, $\pi R^2$ with $R = 0.025$ m, not the area of the coil. $M = \mu_0 nN_2\pi R^2 = 4\pi \times 10^{-7}\ \text{T m/A} \times 650\ \text{m}^{-1} \times 90 \times \pi \times (0.025\ \text{m})^2 = 1.44 \times 10^{-4}\ \text{H}$. That is $144$ µH.