Check the picture below.
so simply let's get the volume of the non-filled cylinder and add the glass weight.
[tex]\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=1.5\\ h=5 \end{cases}\implies V=\pi (1.5)^2h\implies V=11.25\pi ~in^3 \\\\\\ \stackrel{\textit{weight of the water}}{11.25\pi \cdot 0.6}+\stackrel{\textit{glass weight}}{1.06}~\approx~\underset{oz}{36.40}[/tex]
