A silicon crystal is doped with 5 × 10^ 17 \ cm^ -3 atoms of boron....

Boron is a trivalent impurity, so doping silicon with boron creates a p-type semiconductor.

The hole concentration is approximately equal to the acceptor concentration:

$n_h \approx N_A = 5 \times 10^{17}\ \text{cm}^{-3}$

According to the mass action law:

$n_e n_h = n_i^2$

Substituting the given values:

$n_e = \dfrac{n_i^2}{n_h}$

$n_e = \dfrac{(1.5 \times 10^{10})^2}{5 \times 10^{17}}$

$n_e = \dfrac{2.25 \times 10^{20}}{5 \times 10^{17}}$

$n_e = 0.45 \times 10^3\ \text{cm}^{-3}$

$n_e = 4.5 \times 10^2\ \text{cm}^{-3}$

Answer: $4.5 \times 10^2\ \text{cm}^{-3}$