Answer
$7.2\cdot10^{-4}$ H
Work Step by Step
At position $a$, we can assume that the switch has been closed for a long time which means that there is a steady current in the circuit (a current that is not changing). We can find the current at this instant by using Kirchoff's law:
$\varepsilon - IR - \displaystyle L\frac{dI}{dt} = 0$
$3 - I(50) - L(0) = 0\qquad\rightarrow\qquad3-I(50)=0$
$\displaystyle I = \frac{\varepsilon}{R} = \frac{3}{50}$ Amps
As soon as the switch is moved from position $a$ to position $b$, the steady current in the system at position $a$ is the initial current as soon as we move to position $a$.
$I_0 = \displaystyle \frac{\varepsilon}{R}= \frac{3}{50}$ Amps
Given that after $5$ microseconds the energy in the inductor is half of the initial energy, and knowing that $U_L = \displaystyle \frac{1}{2}LI^2$ and for $LR$ circuits $I(t) = I_0e^{-t/(L/R)}$ we can solve for the inductance of the inductor.
$\displaystyle \frac{1}{2} = \frac{U_f}{U_0} = \frac{\frac{1}{2}LI_f^2}{\frac{1}{2}LI_0^2} = \frac{I_f^2}{I_0^2} = \frac{(I_0e^{-t/(L/R)})^2}{(\varepsilon/R)^2}$
$\displaystyle \frac{(\varepsilon/R)^2}{2} = (I_0e^{-t/(L/R)})^2$
$\displaystyle \sqrt{\frac{(\varepsilon/R)^2}{2}} = I_0e^{-t/(L/R)}$
$\displaystyle \frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0} = e^{-t/(L/R)}$
$\displaystyle \ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0}) = -t/(L/R) = \frac{-t}{L/R} = \frac{-Rt}{L}$
$\displaystyle \frac{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})}{-Rt} = \frac{1}{L}$
$\displaystyle \frac{-Rt}{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})} = L$
After finding this awful looking equation, we see that our answer is
$L = \displaystyle \frac{-Rt}{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})} = \displaystyle \frac{-(50)(5\cdot10^{-6})}{\ln(\frac{\sqrt{\frac{(3/50)^2}{2}}}{(3/50)})} = 7.2\cdot10^{-4}$ H
