Derivation of Fresnel distance An aperture of size $a$ illuminated by a parallel beam sends diffracted beam (the central maximum) in angular width approximately $\lambda/a$. Travelling a distance $z$, it acquires the width $z \lambda / a$ due to diffraction. The distance at which this width equals the size of the aperture is called the Fresnel distance.
As far as I know, the angular width of central maximum is given by $2 \lambda / a$ and not $\lambda/a$.
 A: This is really just a geometry problem since we can not use Fraunhofer approximation in general. The solution depends on what you mean by "width" of the central maximum. If you take it to be the distance between the two adjacent minima (full width) then the solution is:
$$z_{\text{Fresnel}} = \frac{a^2}{4 \lambda} - \frac{\lambda}{4}$$
Alternatively the full width half maximum would probably have a factor of 2 in front or something like that. Note that for $\lambda \ll a$ the result becomes what LC7 wrote in his answer (who calculated full width half maximum), but it is not the most general form.
Here is my working. The crucial step is that at the minimum the interference is fully destructive, so for every point in the aperture exists another point in the aperture that is $\pi$ out of phase. Hence the phase difference at the minimum between edge of the aperture and the centre of the aperture is $\pi$ (see picture). The rest is then just doing the geometry.

EDIT in response to comments
I will try to give an explanation of how this fits with the question.

The distance at which this width equals the size of the aperture is called the Fresnel distance.

I calculated this distance above, fixing the ambiguous "width" to be the full width, as I explained (as opposed to full width half maximum). My derivation does not make any approximation other than scalar diffraction theory.

An aperture of size $a$ illuminated by a parallel beam sends diffracted beam (the central maximum) in angular width approximately $\lambda/a$. Travelling a distance $z$, it acquires the width $z \lambda / a$ due to diffraction.

The OP talks about "beams" here. This is only valid in the Fraunhofer approximation, i.e. the far field limit. Why? Interference/diffraction causes a complicated wave field behind the aperture. Far away from it it looks a bit simpler, since only the linear phase terms will be significant. Now linear phase terms give you plane waves. These are the "beams" you are talking about. However the Fresnel distance does not lie in the Frauhofer regime region!! So unless you make further approximations this is not a valid approach.

As far as I know, the angular width of central maximum is given by $2 \lambda / a$ and not $\lambda/a$.

Again this is a property of the diffraction pattern in the Fraunhofer limit and not true in general, which is where the whole misconception arises.
Comment about the comments
From what I gather from the comments people are just trying to find a derivation of some particular formula they learnt in some course. If that is the case I think the question should be closed. I think I have explained all conceptual aspects of the problem and am open to further suggestions about how to improve that, but I won't go into throwing around formulae without concepts.
A: "As far as I know, the angular width of central maximum is given by 2λ/a and not λ/a."
The half-power angular width of the main beam is λ/a radians. That's why it is preferred.
A: It is $\frac{\lambda}a$  because we are talking about only two rays, so the formula of double slit fringe width is more appropriate.  If there were a vast no. of rays, $\frac{2\lambda}d$ would be correct.  As told by Feynman the only significant difference between diffraction and interference is the number of light sources at slit.
A: $I(\theta)= I_{max} \left[ \dfrac{\sin\left(\frac{\pi a \sin \theta}{\lambda}\right)}{\frac{\pi a \sin \theta}{\lambda}}\right]^2$
where $a$ is the size of aperture.
This is the diffraction pattern, now you can easily see that the minimum of diffraction are where $\sin \theta = m \lambda /a$. So as you say the angular width of central maxima is $2 \lambda /a$, in fact from the relation $\Delta x = z \Delta \theta$ with $z$ travelling distance and $\Delta x$ width of the central maximum on the detector, in approximation $a\gg \lambda \Rightarrow \Delta \sin \theta \simeq \Delta \theta = 2 \lambda /a$.
We obtain that $\Delta x=z 2 \lambda /a$. Now the Fresnel distance is where $\Delta x = a$ i.e.: 
$a=2 z \lambda /a \Rightarrow  z_{fresnel}= \frac {a^2}{2 \lambda}$
That is the final result.
