A small coil of 30 turns has an area of 8.0 cm² for each turn…

Step 1. The change of flux. $\Delta\Phi = 2BA = 2 \times 0.50 \times 10^{-3}\ \text{T} \times 8.0 \times 10^{-4}\ \text{m}^2 = 8.0 \times 10^{-7}\ \text{Wb}$.

Step 2. The charge. $Q = \dfrac{N\Delta\Phi}{R} = \dfrac{30 \times 8.0 \times 10^{-7}\ \text{Wb}}{6.0\ \Omega} = 4.0 \times 10^{-6}\ \text{C}$, that is $4.0\ \mu$C.

Step 3. The average emf. $\varepsilon = \dfrac{N\Delta\Phi}{\Delta t} = \dfrac{30 \times 8.0 \times 10^{-7}\ \text{Wb}}{0.10\ \text{s}} = 2.4 \times 10^{-4}\ \text{V}$, that is $0.24$ mV.

Step 4. The other place. We want $B$, so $B = \dfrac{QR}{2NA} = \dfrac{8.0 \times 10^{-6}\ \text{C} \times 6.0\ \Omega}{2 \times 30 \times 8.0 \times 10^{-4}\ \text{m}^2} = 1.00 \times 10^{-3}\ \text{T}$, that is $1.00$ mT.