The speed at the top can be found from the conservation of energy for the bead-track-Earth system,
We define the bottom of the loop as the zero level for the gravitational potential energy.

$\mathrm{Since} {\mathrm{v}}_{\mathrm{i}} = 0, {\mathrm{E}}_{\mathrm{i}} = {\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{i}} = 0 + \mathrm{mgh} = \mathrm{mg}(3.50\mathrm{R})$

The. total energy of the bead at point (A) can be written as ${\mathrm{E}}_{\mathrm{A}} = {\mathrm{K}}_{\mathrm{A}}+{\mathrm{U}}_{\mathrm{A}} = \frac{1}{2}{{\mathrm{mv}}_{\mathrm{A}}}^2+\mathrm{mg}\left(2\mathrm{R}\right)$

Since mechanical energy is conserved, ${\mathrm{E}}_{\mathrm{i}} = {\mathrm{E}}_{\mathrm{A}}, \mathrm{we} \mathrm{get}$

$\mathrm{mg}(3.50\mathrm{R}) = \frac{1}{2}{{\mathrm{mv}}_{\mathrm{A}}}^2+\mathrm{mg}\left(2\mathrm{R}\right)$

simpliflying, ${{\mathrm{v}}_{\mathrm{A}}}^2 = 3.0 \mathrm{gR} \Rightarrow {\mathrm{v}}_{\mathrm{A}} = \sqrt{3.0 \mathrm{gR}}$

To find the normal force at the top, we construct a force diagram as shown,

Here we assume that N is downward, like mg.

$\sum _{ }{\mathrm{F}}_{\mathrm{y}} = {\mathrm{ma}}_{\mathrm{y}}; \mathrm{N} + \mathrm{mg} = \frac{{\mathrm{mv}}^2}{\mathrm{r}}$

$\mathrm{N} = \mathrm{m}[\frac{{\mathrm{v}}^2}{\mathrm{R}}-\mathrm{g}] = \mathrm{m}[\frac{3.0\mathrm{gR}}{\mathrm{R}}-\mathrm{g}] = 2.0 \mathrm{mg}$

$\mathrm{N} = 2.0(50\times {10}^{-3}\mathrm{kg})(10.0 \mathrm{m}/{\mathrm{s}}^2)$

   = 1.0 N downward