A solenoid 0.30 m long and 1.2 cm in radius must have an inductance of 25 mH…

Step 1. The area. $A = \pi r^2 = \pi \times (0.012\ \text{m})^2 = 4.52 \times 10^{-4}\ \text{m}^2$.

Step 2. The air core. $N = \sqrt{\dfrac{Ll}{\mu_0 A}} = \sqrt{\dfrac{25 \times 10^{-3}\ \text{H} \times 0.30\ \text{m}}{4\pi \times 10^{-7}\ \text{T m/A} \times 4.52 \times 10^{-4}\ \text{m}^2}} = 3632$ turns.

Step 3. The wire. One turn is $2\pi r = 0.0754$ m long. So the wire is $3632 \times 0.0754\ \text{m} = 274$ m.

Step 4. The iron rod. $L$ is $400$ times larger for the same turns. So the turns fall by $\sqrt{400} = 20$: $N = \dfrac{3632}{20} = 182$ turns.