The magnetic potential energy stored in a certain inductor is 25mJ...

For current& $\text{I}$~through an inductor coil having inductance& $\text{L}$~, Magnetic potential energy stored in the& inductor is given by&

$U = \dfrac{1}{2}LI^{2}$

Given~

${U=25 mJ=25×10}^{- 3}\text{J}$

${I=60 mA =60×10}^{- 3}\text{A}$

$∴ 25 \times {10}^{- 3} = \dfrac{1}{2}L\left(60 \times {10}^{- 3}\right)2$
Solving~$L = 13.89\text{H}$