The hollow cylinder and sphere both are rolling down without slipping therefore the acceleration down the inclined plane for both is, $a_{\mathrm{cylinder}}=\frac{g\sin \theta }{\left(1+{\displaystyle \frac{I_{\mathrm{cylinder}}}{m{r}^2}}\right)}$and $a_{\mathrm{sphere}}=\frac{g\sin 30°}{\left(1+{\displaystyle \frac{I_{\mathrm{sphere}}}{m{r}^2}}\right)}$

For time taken to reach the bottom surface, $s=ut+\frac{1}{2}a{t}^2$$$\begin{gathered} s=\frac{1}{2}a{t}^2 \\ \Rightarrow t=\sqrt{\frac{2s}{a}} \\ \Rightarrow t=\sqrt{\frac{2s}{\left[\frac{g\sin \theta }{\left(1+{\displaystyle \frac{I}{m{r}^2}}\right)}\right]}}=\sqrt{\frac{2s\left(m{r}^2+I\right)}{\left(m{r}^2\right)g\sin \theta }} \end{gathered}$$

For sphere moment of inertia is, $I_{\mathrm{spere}}=\frac{2}{5}m{r}^2$therefore, $$\begin{gathered} t\text{'}=\sqrt{\frac{2s\left(m{r}^2+I\right)}{\left(m{r}^2\right)g\sin 30°}} \\ t\text{'}=\sqrt{\frac{2s\left(1+\frac{2}{5}\right)}{g\sin 30°}}=\sqrt{\frac{14s}{5g\sin 30°}} \\ t\text{'}=\sqrt{\frac{28s}{5g}} \end{gathered}$$

Now for hollow cylinder moment of inertia is, $I_{\mathrm{cylinder}}=m{r}^2$therefore, $$\begin{gathered} t=\sqrt{\frac{2s\left(m{r}^2+I\right)}{\left(m{r}^2\right)g\sin \theta }} \\ t=\sqrt{\frac{2s\left(1+1\right)}{g\sin \theta }}=\sqrt{\frac{4s}{g\sin \theta }} \end{gathered}$$

Given that $t\text{'}=t,$

$$\begin{gathered} \therefore \sqrt{\frac{28s}{5g}}=\sqrt{\frac{4s}{g\sin \theta }} \\ \Rightarrow \sin \theta =\frac{20}{28} \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{5}{7}\right)=45° \end{gathered}$$