Explanation

Maximum current in the RL circuit:

$$ I_0 = \frac{V}{R} = \frac{10}{10} = 1\ \text{A} $$

At the given instant:

$$ I = \frac{I_0}{e} $$

Magnetic field inside the solenoid:

$$ B = \mu_0 n I $$

$$ B = (4\pi \times 10^{-7})(10^4)\left(\frac{1}{e}\right) = \frac{4\pi \times 10^{-3}}{e} $$

Energy density:

$$ u = \frac{B^2}{2\mu_0} $$

$$ u = \frac{(4\pi \times 10^{-3})^2}{2(4\pi \times 10^{-7})}\times\frac{1}{e^2} $$

$$ u = 20\pi \times \frac{1}{e^2}\ \text{J/m}^3 $$

Hence,

$$ \alpha = \boxed{20} $$

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