A

To find the mutual inductance \( M_{21} \) of the two concentric solenoids, we can use the formula for mutual inductance. The mutual inductance is given by the expression:
\[ M_{21} = \frac{N_1 \Phi_{21}}{I_2} \]
where \( \Phi_{21} \) is the magnetic flux linking the inner solenoid per unit current through the outer solenoid.
Step 1: Calculate the magnetic field \( B_2 \) inside the outer solenoid (solenoid 2), which can be expressed as:
\[ B_2 = \mu_0 \frac{N_2}{l} I_2 \]
where \( \mu_0 \) is the permeability of free space.
Step 2: Calculate the flux \( \Phi_{21} \) through solenoid 1 due to the magnetic field from solenoid 2. The area of the inner solenoid is given by \( A_1 = \pi r_1^2 \). Therefore,
\[ \Phi_{21} = B_2 \cdot A_1 = \left(\mu_0 \frac{N_2}{l} I_2\right) \cdot \left(\pi r_1^2\right) \]
Step 3: Substitute \( \Phi_{21} \) back into the expression for mutual inductance:
\[ M_{21} = \frac{N_1 \left( \mu_0 \frac{N_2}{l} I_2 \cdot \pi r_1^2 \right)}{I_2} \]
Cancelling \( I_2 \) from numerator and denominator gives:
\[ M_{21} = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]
This result shows that mutual inductance is directly proportional to the product of the turns of both solenoids and inversely proportional to their length and depends on the cross-sectional area of the inner solenoid.
Therefore, the mutual inductance of the inner solenoid due to the outer solenoid is
\[ M_{21} = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]. Therefore, the correct answer is option A.