Answer
$2\sqrt{7}$
Work Step by Step
Let ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle,\qquad {\bf w}=\langle w_{1},w_{2},w_{3}\rangle$
${\bf v-2w}=\langle v_{1}-2w_{1},v_{2}-2w_{2},v_{3}-2w_{3}\rangle$
$|{\bf v-2w}|^{2}=(v_{1}-2w_{1})^{2}+(v_{2}-2w_{2})^{2}+(v_{3}-2w_{3})^{2}$
$=(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})+4(w_{1}^{2}+w_{2}^{2}+w_{3}^{2})-4(v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3})$
$=|{\bf v}|^{2}+4|{\bf w}|^{2}-4({\bf v}\cdot{\bf w})$
We are given $|{\bf v}|, |{\bf w}|$ and we know that
${\bf v}\cdot{\bf w}=|{\bf v}|\cdot |{\bf w}|\cdot\cos\theta\qquad $
$|{\bf v-2w}|^{2}=2^{2}+4\displaystyle \cdot 3^{2}-4\cdot 2\cdot 3\cdot\cos\frac{\pi}{3}$
$|{\bf v-2w}|^{2}=4+36-12=28$
$|{\bf v-2w}|\}=\sqrt{28}=2\sqrt{7}$
