Answer
$9.35\cdot10^{-2}$ A
Work Step by Step
The current in an $LR$ circuit can be expressed as $I(t) = I_0e^{-t/\tau}$ where $\tau$ is the time constant of the circuit and $\tau = \displaystyle \frac{L}{R}$
We have to make a key assumption in this problem that when the circuit is in position "a" that the switch has been there for a long time and that there is a steady current in the circuit such that $\displaystyle \frac{dI}{dt} = 0$. Doing so and using Kirchoff's loop law, we find that
$\varepsilon - IR - L(0) = 0$
$I = \displaystyle \frac{\varepsilon}{R} = I_0$
When the switch is flipped to position "b", our $LR$ is created and we can now say that the current can be expressed as $I(t) = \displaystyle \frac{\varepsilon}{R}e^{-t/(L/R)}$
The problem is that we are not explicitly given the inductance, $L$, to calculate the time constant. Luckily, for us, however, we are super smart and remember that in chapter 30.8 the inductance of a solenoid is given as $L_{solenoid} = \displaystyle \frac{\mu_0N^2A}{l}$
The inductance of the inductor (which in this case is a solenoid) is then $L_{inductor} = \displaystyle \frac{4\pi\cdot10^{-7}(300)^2(\pi(0.01)^2)}{0.09} \approx 3.95\cdot10^{-4}$ H
Now, the current in the circuit at $t = 10\cdot10^{-6} = 10^{-5}$ seconds is
$I(10^{-5}) = \displaystyle \frac{6.0}{30}e^{-10^{-5}/(3.95\cdot10^{-4}/30)} = 9.35\cdot10^{-2}$ A
