Which atom is it?

The total charge of the atomic nucleus,

$q=\int { \rho (r)dV } = \int_0 ^R { \rho_0 \left[ 1 - \left( \dfrac {r}{R} \right) ^2 \right] (4 \pi r^2) dr }$

$\Rightarrow q = 4 \pi \rho_0 \int_0 ^R { \left[ r^2 - \dfrac {r^4}{R^2} \right] dr} = 4 \pi \rho_0 \left[ \dfrac{r^3}{3} - \dfrac{r^5}{5R^2} \right] _0^ R = 4 \pi \rho_0 \left[ \dfrac{r^3}{3} - \dfrac{r^3}{5} \right]$

$\quad \quad = \dfrac {8 \pi \rho_0 R^3} {15} = \frac{8}{15} \pi \times 5 \times 10^{25} \times (3.4 \times 10^{-15})^3 = 3.29272 \times 10^{-18} C$

The electronic charge $e = 1.60217657 \times 10^{-19} C$ , therefore,

$q = \dfrac {3.29272 \times 10^{-18}} {1.60217657 \times 10^{-19}} = 20.55 \approx \boxed {21}$ electrons (scandium).