Answer
$U_i=5.63\times10^{-5}J$, and 0.537 s.
Work Step by Step
Refer to section 21-11 to find the initial energy stored in the inductor.
$$U_i=\frac{1}{2}LI^2=\frac{1}{2}(0.0450H)(0.0500A)^2=5.63\times10^{-5}J$$
If the energy increases by a factor of 5, since it is proportional to the square of the current, the final current is $\sqrt{5}$ times greater than the initial current.
$$U_f=5U_i$$
$$I_f=\sqrt{5}I_i$$
The current increases at a constant rate, so find the time.
$$\frac{\Delta I}{\Delta t}=0.115A/s = \frac{I_f-I_i}{\Delta t}$$
$$\Delta t =\frac{I_f-I_i}{0.115A/s }=\frac{\sqrt{5}I_i -I_i}{0.115A/s }$$
$$\Delta t =\frac{(\sqrt{5}-1)I_i }{0.115A/s }$$
$$\Delta t =\frac{(\sqrt{5}-1)0.050A }{0.115A/s }=0.537s$$