Answer
See the detailed answer below.
Work Step by Step
Since the string vibrates at its second-harmonic frequency, the wavelength of the standing wave is then equal to the length of the string; $(\lambda= L)$.
We know that the displacement of a medium in this situation is given by
$$D_{(x,t)}=A_{(x)} \cos(\omega t)$$
where
$$A_{(x)}=2a\sin(kx)=A_{max}\sin(kx)$$
where $k=2\pi/\lambda$, so
$$A_{(x)}= A_{max}\sin\left(\dfrac{2\pi x}{\lambda}\right)$$
Plugging the known;
$$A_{(x)}= 2.0\sin\left(\dfrac{2\pi x}{2}\right)$$
$$A_{(x)}=2.0\sin\left( \pi x\right)$$
At $x=10$ cm:
$$A_{(x)}=2.0\sin\left( \pi [0.10]\right)=\color{red}{\bf 0.62}\;\rm cm$$
At $x=20$ cm:
$$A_{(x)}=2.0\sin\left( \pi [0.20]\right)=\color{red}{\bf 1.2}\;\rm cm$$
At $x=30$ cm:
$$A_{(x)}=2.0\sin\left( \pi [0.30]\right)=\color{red}{\bf 1.6}\;\rm cm$$
At $x=40$ cm:
$$A_{(x)}=2.0\sin\left( \pi [0.40]\right)=\color{red}{\bf 1.9}\;\rm cm$$
At $x=50$ cm:
$$A_{(x)}=2.0\sin\left( \pi [0.50]\right)=\color{red}{\bf 2.0}\;\rm cm$$
