The correct option is B energy density
Energy stored = 12LI2 L=inductance of solenoid
μ0N2Al
Energy Density = 12LI2AL l=lengthofthesolenoidA=crossectionareaN=noofturnsB=Magneticfield=\dfrac{\dfrac{1}{2}\dfrac{\mu _{0}N^{2}A}{l}\times \dfrac{B^{2}l^{2}}{\mu {_{0}}^{2}N^{2}}}{Al}=\dfrac{1}{2}\dfrac{B^{2}}{\mu _{0}}$