Let $\mathrm{h}$ be the maximum height attained and $\mathrm{v}$ be the velocity at maximum height. From conservation of angular momentum

$\mathrm{m}\frac{{\mathrm{V}}_{\mathrm{e}}}{2}\cos 45\times \mathrm{R}=\mathrm{mvr}$ $($where $\left.r=R+h\right)$

$\frac{{\mathrm{V}}_{\mathrm{e}}}{2}\times \frac{1}{\sqrt{2}}\mathrm{R}=\mathrm{vr}\Rightarrow \mathrm{v}=\frac{{\mathrm{V}}_{\mathrm{e}}}{2\sqrt{2}}\frac{\mathrm{R}}{\mathrm{r}}=\frac{1}{2\sqrt{2}}\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}\times \frac{\mathrm{R}}{\mathrm{r}}\Rightarrow \mathrm{v}=\sqrt{\frac{\mathrm{GMR}}{4{\mathrm{r}}^2}}$

From conservation of energy

$\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2}\mathrm{m}\frac{{\mathrm{V}}_{\mathrm{o}}^2}{4}=-\frac{\mathrm{GMm}}{\mathrm{r}}+\frac{1}{2}{\mathrm{mv}}^2\Rightarrow \frac{-\mathrm{GM}}{\mathrm{R}}+\frac{1}{2}\times \frac{1}{4}\times 2\frac{\mathrm{GM}}{\mathrm{R}}=-\frac{\mathrm{GM}}{\mathrm{r}}+\frac{1}{2}\times \frac{\mathrm{GMR}}{4{\mathrm{r}}^2}$

$-\frac{3}{4\mathrm{R}}=\frac{-8\mathrm{r}+\mathrm{R}}{8{\mathrm{r}}^2}\Rightarrow -6{\mathrm{r}}^2=-8\mathrm{rR}+{\mathrm{R}}^2\Rightarrow 6{\mathrm{r}}^2-8\mathrm{rR}+{\mathrm{R}}^2=0$

$\mathrm{r}=\frac{8\mathrm{R}\pm \sqrt{64{\mathrm{R}}^2-4\times 6\times {\mathrm{R}}^2}}{12}\Rightarrow \mathrm{r}=\frac{8\mathrm{R}\pm 2\sqrt{10}\mathrm{R}}{12}\Rightarrow \mathrm{R}+\mathrm{h}=\frac{4\mathrm{R}+\sqrt{10}\mathrm{R}}{6}\Rightarrow \mathrm{h}=0.19\mathrm{R}$