We know that, the internal bisector divides base in the same ratio as the two remaining sides.

$\Rightarrow \frac{OA}{OB}=\frac{AC}{BC}$

$\Rightarrow \frac{\sqrt{2^2+2^2+1^2}}{\sqrt{2^2+4^2+4^2}}=\frac{AC}{BC}$

$\Rightarrow \frac{3}{6}=\frac{AC}{BC}$

$\Rightarrow AC:BC=1:2$

$\Rightarrow C\equiv \left(\frac{1\times 2+2\times 2}{3},\frac{1\times 4+2\times 2}{3},\frac{1\times 4+2\times 1}{3}\right)$

$\Rightarrow C\equiv \left(2,\frac{8}{3},2\right)$

$\Rightarrow OC=\sqrt{2^2+{\left(\frac{8}{3}\right)}^2+2^2}$

$\Rightarrow OC=\sqrt{8+\frac{64}{9}}=\frac{\sqrt{136}}{3}$

$\Rightarrow OC=\frac{2\sqrt{34}}{3}$