$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left(1-{\mathrm{e}}^{-\frac{\mathrm{Rt}}{\mathrm{L}}}\right)=5\left(1-{\mathrm{e}}^{-2\mathrm{t}}\right)$

When potential difference across R is 4 V, the current is

$\begin{array}{l} \mathrm{I}=\frac{4}{4}=1\mathrm{A}\\ \Rightarrow 1=5\left(1-{\mathrm{e}}^{-2\mathrm{t}}\right)\\ \Rightarrow {\mathrm{e}}^{-2\mathrm{t}}=0.8\end{array}$

At this instant, we have

$\frac{\mathrm{dI}}{\mathrm{dt}}=10{\mathrm{e}}^{-2\mathrm{t}}=100\times 0.8=8{\mathrm{As}}^{-1}$

Power supplied by battery is

$\mathrm{P}=\mathrm{EI}=20\times 1=20\mathrm{W}$

Power consumed by resistor is

${\mathrm{P}}_{\mathrm{R}}={\mathrm{I}}^2\mathrm{R}=1^2\times 4=4\mathrm{W}$

So, power stored in magnetic field is

${\mathrm{P}}_{\mathrm{m}}=20\mathrm{W}-4\mathrm{W}=16\mathrm{W}$