A long air-core solenoid has 2500 turns, a length of 0.80 m and a radius of 3.0 cm…

Step 1. The field. $n = \dfrac{2500}{0.80\ \text{m}} = 3125\ \text{m}^{-1}$. $B = \mu_0 nI = 4\pi \times 10^{-7}\ \text{T m/A} \times 3125\ \text{m}^{-1} \times 4.0\ \text{A} = 1.57 \times 10^{-2}\ \text{T}$. That is $15.7$ mT.

Step 2. The energy of the coil. $A = \pi r^2 = 2.83 \times 10^{-3}\ \text{m}^2$. $L = \dfrac{\mu_0N^2A}{l} = 0.0278\ \text{H}$. $U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \times 0.0278\ \text{H} \times (4.0\ \text{A})^2 = 0.222\ \text{J}$.

Step 3. The energy in each cubic metre. $u = \dfrac{B^2}{2\mu_0} = \dfrac{(1.57 \times 10^{-2}\ \text{T})^2}{2 \times 4\pi \times 10^{-7}\ \text{T m/A}} = 98.2\ \text{J/m}^3$.

Step 4. The same energy. The volume is $Al = 2.83 \times 10^{-3}\ \text{m}^2 \times 0.80\ \text{m} = 2.26 \times 10^{-3}\ \text{m}^3$. Then $u \times Al = 98.2\ \text{J/m}^3 \times 2.26 \times 10^{-3}\ \text{m}^3 = 0.222\ \text{J}$. It is the same.