The smallest value of k, for which both the roots of the equation x² - kx + 16(k² - k + 1) = 0 are real, distinct and have values at least 4 is

x² - kx + 16(k² - k + 1) = 0
As the roots are real and distinct, we have D > 0 ⇒ 64k² - 64(16(k² - k + 1)) > 0 ⇒ k² - 16(k² - k + 1) > 0 ⇒ k² - 16k² + 16k - 16 > 0 ⇒ -15k² + 16k - 16 > 0 ⇒ 15k² - 16k + 16 < 0. The discriminant of this quadratic is 16² - 4(15)(16) = 256 - 960 = -704 < 0. Since the parabola opens upwards, this inequality has no real solution. However, we must have k>1 (from the condition that the discriminant is positive).
As the roots have values at least 4, we have -b/2a ≥ 4 ⇒ k/2 ≥ 4 ⇒ k ≥ 8.
Let f(x) = x² - kx + 16(k² - k + 1). Since both roots are at least 4, f(4) ≥ 0 ⇒ 16 - 4k + 16(k² - k + 1) ≥ 0 ⇒ 16 - 4k + 16k² - 16k + 16 ≥ 0 ⇒ 16k² - 20k + 32 ≥ 0 ⇒ 4k² - 5k + 8 ≥ 0. The discriminant is 25 - 4(4)(8) = 25 - 128 = -103 < 0. Since the parabola opens upwards, this inequality is always true for any real k.
Combining the conditions, we have k ≥ 8. Therefore, the smallest value of k is 8.