Answer
$I=\frac{1}{2}ma^2$
Work Step by Step
We can find the required moment of inertia as follows:
$I=\Sigma (I^{\prime}+md^2)$
$\implies I=3(\frac{ma^2}{12}+m(\frac{\sqrt{3}}{6}a)^2$
$\implies I=3(\frac{ma^2}{12}+m\frac{3}{36}a^2)$
$\implies I=\frac{1}{2}ma^2$