A particle moves along a circle of radius $\frac{20}{\pi}$ with constant tangential acceleration. 
If the speed of the particle is $80\frac{\text{m}}{\text{s}}$ at the end of second revolution. Find the tangential acceleration (object starts from rest)

$V=R\omega$
$\omega =4\pi$ rad/s
Now 
$$\omega^2=2\alpha\theta$$
$$16\pi^2=8\pi\alpha$$
$$\alpha=2\pi $$
Then 
$$a=R\alpha$$
$$a=40m/s^2$$
But the answer is $80.$ What’s going wrong?
To be clear, I have a perfectly valid solution with me that is convincing enough to get 80m/s^2. I just want to know what’s wrong with this
Upon request, here is the solution 
 A: Your solution seems correct, and the book's solution is wrong. First of all, $d = R\theta = 4\pi R.$ Look at the first 2 equalities after the "$\therefore$":

$a = \frac{v^2}{ad} = \frac{v^2}{2(2\pi R)}$

The first equality has a typo; it should be $\frac{v^2 }{2d}$. Assuming this, $\frac{v^2}{2d} =  \frac{v^2}{2(2\pi R)}$ so $d =  2\pi R \neq 4\pi R, \quad\Rightarrow\!\Leftarrow.$ 
A: Your answer of $a = 40$ m/s$^2$ is correct. Assuming $2$ revolutions, the proposed solution has an arithmetic mistake. The first line is correct, but the second line is wrong. It should be
\begin{align}
a &= \dfrac{v^2}{2d} = \dfrac{v^2}{2 (2 \cdot (2 \pi R))} = 40
\end{align}
First equality is by rearranging $v = \sqrt{2ad}$. The second is by substituting $d = 2(2 \pi R)$, which is the distance travelled at the end of two revolutions and the third is by substituting $v=80$ at the end of two cycles, $R= 20/\pi$.
So, the mistake is that the solution is missing an extra factor of $2$ in the denominator.

(of course, insert units everywhere to be formally correct)
