The smallest integral value of $k$ for which both the roots of $x^2 - 8kx + 16(k^2 - k + 1) = 0$ are real and distinct and have value at least $4,$ is 

Given equation is  $x^2-8kx+16\left(k^2-k+1\right)=0$

Since the roots are real and distinct, so 

$\begin{array}{l} 64k^2-64\left(k^2-k+1\right)>0\\ \Rightarrow k^2-\left(k^2-k+1\right)>0\\ \Rightarrow k-1>0\\ \Rightarrow k>1\end{array}$

Thus, $k=2, 3$

Hence the smallest integral value of $k$ is $2.$