1. Given quadratic

   $x^2-8kx+16(k^2-k+1)=0$

   Let its roots be $\alpha,\beta$.

2. Conditions on the roots

   We need both roots to be:

   • real,
   • distinct,
   • and each at least $4$.

3. Make the condition "roots at least 4" easier

   Put $x=y+4$ so that requiring $x\ge 4$ becomes requiring $y\ge 0$.

   Substitute $x=y+4$ into the equation:

   $(y+4)^2-8k(y+4)+16(k^2-k+1)=0$

   Expand: $y^2+8y+16-8ky-32k+16k^2-16k+16=0$

   $y^2+(8-8k)y+(16k^2-48k+32)=0$

   Factor constants: $y^2+8(1-k)y+16(k^2-3k+2)=0$

   $y^2+8(1-k)y+16(k-1)(k-2)=0$

4. For both roots in $x$ to be at least 4, both roots in $y$ must be non-negative

   Let roots of this new equation be $y_1,y_2$. Then we need:

   • $y_1,y_2$ real and distinct,
   • $y_1\ge 0,\ y_2\ge 0$.

   For a quadratic with leading coefficient $1$, this requires:

   • discriminant $>0$,
   • sum of roots $\ge 0$,
   • product of roots $\ge 0$.

5. Compute these quantities

   For $y^2+8(1-k)y+16(k-1)(k-2)=0,$

   we have $y_1+y_2=-8(1-k)=8(k-1),$ $y_1y_2=16(k-1)(k-2).$

   So,

   (i) Sum non-negative: $8(k-1)\ge 0 \implies k\ge 1.$

   (ii) Product non-negative: $16(k-1)(k-2)\ge 0 \implies (k-1)(k-2)\ge 0.$ Hence, $k\le 1 \quad \text{or} \quad k\ge 2.$

   Combining with $k\ge 1$, we get $k=1 \quad \text{or} \quad k\ge 2.$

6. Distinct real roots condition

   Discriminant of the original quadratic: $\Delta = (-8k)^2-4\cdot 1\cdot 16(k^2-k+1)$

   $\Delta =64k^2-64(k^2-k+1)$

   $\Delta =64(k-1).$

   For real and distinct roots: $\Delta>0 \implies k-1>0 \implies k>1.$

7. Combine all conditions

   From step 5: $k=1$ or $k\ge 2$.

   From step 6: $k>1$.

   Therefore, $k\ge 2.$

   So the smallest value of $k$ is $\boxed{2}.$

8. Verification at $k=2$

   Substitute $k=2$: $x^2-16x+16(4-2+1)=0$ $x^2-16x+48=0$ $(x-4)(x-12)=0$

   Roots are $4$ and $12$, which are real, distinct, and both at least $4$.

Hence the required smallest value is $2$.