1. Use Bohr’s quantization condition for the first orbit

For the electron in the first Bohr orbit $(n=1)$, $mvr=\frac{nh}{2\pi}=\frac{h}{2\pi}$

So, $v=\frac{h}{2\pi mr}$

2. Write centripetal acceleration

The electron moves in a circular orbit, so its acceleration is centripetal: $a=\frac{v^2}{r}$

Substitute $v$ from above: $a=\frac{1}{r}\left(\frac{h}{2\pi mr}\right)^2$

3. Simplify

$a=\frac{1}{r}\cdot \frac{h^2}{4\pi^2 m^2 r^2}$

$a=\frac{h^2}{4\pi^2 m^2 r^3}$

4. Match with the options

This matches: $\boxed{\frac{h^2}{4\pi^2 m^2 r^3}}$

So the correct option is C.