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

$U=\frac{1}{2}L{I}^2$

Given 

${\text{U=25 mJ=25×10}}^{-3}\text{ J}$

${\text{I=60 mA =60×10}}^{-3}\text{ A}$

$\therefore 25\times {10}^{-3}=\frac{1}{2}\mathrm{L}{\left(60\times {10}^{-3}\right)}^2$
Solving $L=13.89 \text{H}$