Answer
$$\sqrt {\dfrac{2}{3}} (\vec{i}-\vec{j}+2\vec{k})$$
Work Step by Step
The unit vector in the direction of $\vec{a}$ can be found as:
$\hat{a}=\dfrac{\vec{a}}{|\vec{a}|} =\dfrac{i-j+2k}{\sqrt 6}$
The unit vector having length $2$ in the direction of $\vec{a}$ can be found as:
$2 \times (\dfrac{i-j+2k}{\sqrt 6})=\sqrt {\dfrac{4}{6}} (\vec{i}-\vec{j}+2\vec{k})\\=\sqrt {\dfrac{2}{3}} (\vec{i}-\vec{j}+2\vec{k})$
