power absorbed in an inductor in an RL circuit during step response
The inductor stores energy when a current passes through it. I don't want to carry around the fact that both current and energy are functions of time, so let the instantaneous energy stored be \$\small E(t)=E\$, when the instantaneous current is \$\small i(t)=i\$. Then we may write \$\small E=\frac{1}{2}Li^2\$. 

Now, the current in your circuit is exponentially increasing, and given by: $$\small i=I(1-e^{-t/\tau})$$ where \$\small I\$ is the maximum current, and \$\small\tau\$ is the time constant.

Hence, the energy at time, \$\small t\$, may be written:$$\small E=\frac{1}{2}Li^2=\frac{1}{2}LI^2(1-e^{-t/\tau})^2=\frac{1}{2}LI^2(1-2e^{-t/\tau}+e^{-2t/\tau})$$
Power, \$\small P=P(t)\$, is the derivative of energy, thus: $$\small P=\frac{dE}{dt}= \frac{LI^2}{\tau}(e^{-t/\tau}-e^{-2t/\tau})$$

This is the equation of the graph of power vs time shown in your question.

Let's check a few important features of the graph against those predicted by the equation:


The equation gives \$\small P(0)=0\$, and \$\small P(\infty)=0\$
Maximum power occurs when: \$\small \frac{dP}{dt}=0\$. Performing this differentiation and ignoring the constant term gives: \$\small (-e^{-t/\tau}+2e^{-2t/\tau})=0\$, hence \$\small \frac{t}{\tau} = 0.693\$. From the circuit, \$\small \tau=\frac{L}{R}=1ms\$, so the peak power is at \$\small t=0.693ms\$
Given the maximum current, \$\small I=10mA\$ and inductance, \$\small L=1H\$, the value of peak power given by the equation at \$\small \frac{t}{\tau}=0.693\$ is \$\small P=25mW\$
When the current has reached its maximum value, the energy stored in the inductance will also have reached its maximum value (\$\small \frac{1}{2}LI^2\$), and the power transferred to the inductance will, henceforth, be zero.
Power and energy transfer to the inductor only takes place when the current through the inductor is changing (i.e. work needs to be done, against the back-emf, to establish a current through the inductor)