Electric field due to given charge distribution is

$E=-\frac{dV}{dr}=-\frac{d}{dr}\frac{q{e}^{-\alpha r}}{4\pi {\epsilon }_{0}r}$

$=\frac{-q}{4\pi {\epsilon }_0}\frac{d}{dr}\left(\frac{e^{-\alpha r}}{r}\right)$

$=\frac{-q}{4\pi {\epsilon }_0}\left(\frac{-\alpha r{e}^{-\alpha r}-e^{-\alpha r}}{r^2}\right)$

$=\frac{q}{4\pi {\epsilon }_0}·e^{-\alpha r}·\left(\frac{\alpha r+1}{r^2}\right)$

Electric field at $r=\frac{1}{\alpha }$ is

$E\left(r=\frac{1}{\alpha }\right)=\frac{q}{4\pi {\epsilon }_0}·e^{-\alpha \times \frac{1}{\alpha }}\left(\frac{\alpha \times \frac{1}{\alpha }+1}{1/{\alpha }^2}\right)$

$=\frac{\left(q/e\right)}{4\pi {\epsilon }_0}·2{\alpha }^2$

Flux through a sphere of radius $\frac{1}{\alpha }$ is

$\varphi =\int E·dA$

For spherical distribution,

$E·dA=EdA\cos 0°=EdA$

and $E=$ uniform.

So, we have

$\therefore \varphi =E\int dA$

$=\frac{q/e}{4\pi {\epsilon }_0}·2{\alpha }^2·4\pi {\left(\frac{1}{\alpha }\right)}^2$

or $\varphi =\frac{2q}{{\epsilon }_{0}e}$

From Gauss' law, we have $\varphi =\frac{q_{\mathrm{enclosed} }}{{\epsilon }_0}$

So, charge enclosed in given sphere is $\frac{2q}{e}$