The expression of the magnetic energy stored in a solenoid in terms of magnetic field $B$, cross-sectional area $A$ and length $l$ of the solenoid, is:

$U_B=\frac{1}{2μ_0}B^2Al$

The correct answer is Option (1) → $U_B=\frac{1}{2μ_0}B^2Al$

$\text{Magnetic energy stored in a solenoid: } U = \frac{1}{2} L I^2$

$\text{Inductance of solenoid: } L = \mu_0 \frac{N^2 A}{l}$

$\text{Magnetic field inside solenoid: } B = \mu_0 \frac{N}{l} I \Rightarrow I = \frac{B l}{\mu_0 N}$

$U = \frac{1}{2} \cdot \mu_0 \frac{N^2 A}{l} \cdot \left(\frac{B l}{\mu_0 N}\right)^2$

$U = \frac{1}{2} \cdot \mu_0 \frac{N^2 A}{l} \cdot \frac{B^2 l^2}{\mu_0^2 N^2}$

$U = \frac{B^2}{2 \mu_0} \cdot A \cdot l$

$\text{Answer: } U = \frac{B^2 A l}{2 \mu_0}$