Let the axis of the solenoid is z axis the change of magnetic field inside the solenoid $\frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{\alpha }$ (constant), given we knwo the integral form $\int \overrightarrow{\mathrm{E}}·\overrightarrow{\mathrm{dl}}=\int \frac{\partial \overrightarrow{\mathrm{B}}}{\partial \mathrm{t}}·\overrightarrow{\mathrm{dA}}$ taking a circular loop fo radius a. where we wont to find electric field.

so, $\left|\overrightarrow{\mathrm{E}}\right|\int \mathrm{dl}=\frac{\mathrm{dB}}{\mathrm{dt}}·\int \mathrm{dA}$

$\Rightarrow \mathrm{E}·2\mathrm{\pi a}=\mathrm{\alpha }·{\mathrm{\pi a}}^2$

$\Rightarrow \mathrm{E}=\frac{{\mathrm{\alpha \pi a}}^2}{2\mathrm{\pi a}}$

$\Rightarrow \mathrm{E}=\frac{\mathrm{\alpha a}}{2}$