Answer
$12.03\;\rm\mu m$
Work Step by Step
Before putting the glass piece in front of one slit, the center of the screen was occupied by the central maximum bright fringe.
And after putting the glass piece the center of the screen is occupied by the 10th dark fringe $m=10$.
Recalling that the number of wavelength in a glass piece which has a thickness of $t$ is given by
$$m_{glass}=\dfrac{t}{\lambda_{glass}}$$
where $\lambda_{glass}=\lambda/n$, where $n$ is the index of refraction of the glass.
$$m_{glass}=\dfrac{nt}{\lambda}\tag 1$$
So, the number of wavelength in a air which has the same thickness of $t$ is given by
$$m_{air}=\dfrac{t}{\lambda}\tag 2$$
Since the 10th dark gringe is placed in the same position of the central maximum, the path length has increased by
$$\Delta m=m_{glass}-m_{air}$$
where $\Delta m$ is about 9.5 fringe since there is 9 dark fringes from the first dark fringe to the tenth dark fringe. Plus the distance from the maximum bright fringe to the first dark fringe is one-half fringe.
Hence,
$$9.5=m_{glass}-m_{air}$$
Plugging from (1) and (2);
$$9.5=\dfrac{nt}{\lambda}-\dfrac{ t}{\lambda}=(n-1)\dfrac{t}{\lambda}$$
Thus,
$$t=\dfrac{9.5\lambda}{n-1}$$
Plugging the known;
$$t=\dfrac{9.5(633\times 10^{-9})}{1.5-1}=12.03\times 10^{-6}\;\rm m$$
$$t=\color{red}{\bf12.03}\;\rm\mu m$$
