Đáp số: 4.

\({\rm{V}} = \pi {{\rm{r}}^2}\;{\rm{h}} \Rightarrow {\rm{Sxq}} = 2\pi {\rm{rh}} = 2\pi {\rm{r}} \cdot \frac{{\rm{V}}}{{\pi {{\rm{r}}^2}}} = \frac{{2\;{\rm{V}}}}{{\rm{r}}} \Rightarrow {\rm{Stp}} = \frac{{2\;{\rm{V}}}}{{\rm{r}}} + 2\pi {{\rm{r}}^2} = {\rm{f}}({\rm{r}}).\)

\({{\rm{f}}^\prime }({\rm{r}}) = \frac{{ - 2\;{\rm{V}}}}{{{{\rm{r}}^2}}} + 4\pi {\rm{r}} = \frac{{4\pi {{\rm{r}}^3} - 2\;{\rm{V}}}}{{{{\rm{r}}^2}}} = \frac{{4\pi }}{{{{\rm{r}}^2}}}\left( {{{\rm{r}}^3} - \frac{{\rm{V}}}{{2\pi }}} \right)\)

Lập bảng biến thiên ta có \({\rm{Stp}}\) đạt giá trị nhỏ nhất khi \({\rm{r}} = \sqrt[3]{{\frac{{\rm{V}}}{{2\pi }}}} = 4(\;{\rm{cm}}).\)