The magnetic field $$B_2$$ due to the current $$I_2$$ in the larger coil with 200 turns is given by:

$$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$$

The magnetic flux $$\phi_{1,2}$$ through the smaller coil due to this magnetic field is given by:

$$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$$

Since $$\phi_{1,2} = MI_2$$, we can solve for the mutual inductance $$M$$:

$$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$$

Substituting the given values for $$r_1$$, $$N_1$$, $$r_2$$, and $$N_2$$:

$$M = \frac{10 \times 200 \times 4 \pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 10}$$

Simplifying the expression, we get:

$$M = 4 \times 10^{-8} \mathrm{H}$$

So, the mutual inductance between the two concentric coils is $$4 \times 10^{-8} \mathrm{H}$$.