Answer
$1.41\;\rm cm$
Work Step by Step
Since the string vibrates at its fundamental frequency, the wavelength of the standing wave is then equal to twice the length of the string; $(\lambda=2L)$.
Recalling that the amplitude is given by
$$A_{(x)}=2a\sin(kx)$$
where $k=\frac{2\pi}{\lambda}=\frac{2\pi}{2L}=\frac{\pi}{L}$
$$A_{(x)}=2a\sin\left(\dfrac{\pi x}{L}\right)$$
At $x=\frac{1}{4}L$, $A=2$ cm, so
$$\color{red}{\bf\not} 2=\color{red}{\bf\not} 2a\sin\left(\dfrac{\pi \cdot\frac{1}{4}\color{red}{\bf\not} L}{\color{red}{\bf\not} L}\right)$$
$$a\sin\left(\frac{\pi}{4}\right)=1$$
Hence,
$$a=\dfrac{1}{\sin\left(\frac{\pi}{4}\right)}=\sqrt{2}$$
$$a=\color{red}{\bf1.41}\;\rm cm$$
