Let the refractive index of air, glass, and water be $n_a$, $n_g$, and $n_w$ respectively. The critical angle $c$ is given by the relation $\sin c = \frac{n_2}{n_1}$, where $n_1$ is the refractive index of the denser medium and $n_2$ is the refractive index of the rarer medium. Given, For glass with respect to air, the critical angle $c_g = 42^\circ$. $$ \begin{aligned} \sin 42^\circ &= \frac{n_a}{n_g} \\ \Rightarrow n_g &= \frac{n_a}{\sin 42^\circ} \end{aligned} $$ For water with respect to air, the critical angle $c_w = 48^\circ$. $$ \begin{aligned} \sin 48^\circ &= \frac{n_a}{n_w} \\ \Rightarrow n_w &= \frac{n_a}{\sin 48^\circ} \end{aligned} $$ Now, for the critical angle $c'$ of glass with respect to water, the denser medium is glass and the rarer medium is water. $$ \begin{aligned} \sin c' &= \frac{n_w}{n_g} \\ &= \frac{n_a / \sin 48^\circ}{n_a / \sin 42^\circ} \\ &= \frac{\sin 42^\circ}{\sin 48^\circ} \\ &= \frac{0.6691}{0.7431} \\ &= 0.9004 \end{aligned} $$ Therefore, $$ \begin{aligned} c' &= \sin^{-1}(0.9004) \\ &= 64.22^\circ \end{aligned} $$ The critical angle of glass with respect to water is $64.22^\circ$.