Given data

$M=0.002\,\text{H}, \quad i=i_0\sin \omega t, \quad i_0=5\,\text{A}, \quad \omega=50\pi\,\text{rad/s}$

Formula for induced emf in the second coil

The induced emf due to mutual inductance is $e=-M\frac{di}{dt}$

So we first differentiate the current: $i=5\sin(50\pi t)$ $\frac{di}{dt}=5\cdot 50\pi \cos(50\pi t)=250\pi \cos(50\pi t)$

Expression for emf

$e=-0.002\times 250\pi \cos(50\pi t)$ $e=-0.5\pi \cos(50\pi t)$

Maximum emf

Since maximum value of $|\cos(50\pi t)|$ is $1$, $e_{\max}=0.5\pi=\frac{\pi}{2}\,\text{V}$

Compare with the given form

Given, $e_{\max}=\frac{\pi}{\alpha}$

Therefore, $\frac{\pi}{\alpha}=\frac{\pi}{2}$

Hence, $\alpha=2$

Comparison with stored answer

Derived answer is $2$, which matches the stored correct answer.