Hi divinelogos,

If $\displaystyle \underline{a}\mbox{ and }\underline{b}$ are two vectors the cross product is defined by,

$\displaystyle \underline{a}\times\underline{b}=\underline{\hat{n }}|\underline{a}||\underline{b}|\sin\theta~;~\mbox { where }\theta\mbox{ is the angle between }\underline{a}\mbox{ and }\underline{b}~;~\underline{\hat{n}}\mbox{ is the unit vector perpendicular to }\underline{a}\mbox{ and }\underline{b}$

The direction of $\displaystyle \underline{\hat{n}}$ is given by the

Right hand rule.
Therefore, $\displaystyle e_{1}\times e_{2}=\underline{\hat{n}}|e_{1}||e_{2}|\sin \frac{\pi}{6}=\frac{1}{2}\underline{\hat{n}}$

Since $\displaystyle e_{3}$ is perpendicular to $\displaystyle e_{1}\mbox{ and }e_{2}$;

$\displaystyle \underline{\hat{n}}=e_{3}$

$\displaystyle \therefore e_{1}\times e_{2}=\frac{1}{2}e_{3}$

It should be noted that, we have to assume that $\displaystyle \underline{\hat{n}}\neq -e_{3}$ to get the given answer.