Given, diameter of circle, d = 50 cm
$=50\times {10}^{-2}m$
And frequency, f = 2 Hz
The acceleration of particle in a uniform circular motion can be given as
$a={\omega }^{2}x$
Where, $\omega$ = angular frequency = $2\pi f$
X = distance from centre $=\frac{d}{2}$
$\Rightarrow a=4{\pi }^2{f}^2\times \frac{d}{2}$ … (i)
Substituting given values in Eq. (i), we get
$a=4{\pi }^2\times 4\times \frac{50\times {10}^{-2}}{2}=4{\pi }^2$