Answer
$KE_r$ is greater for axis B.
Work Step by Step
Rotational kinetic energy is calculated by $$KE_r=\frac{1}{2}I\omega^2=\frac{1}{2}\sum MR^2\omega^2$$
In both cases, the total mass $\sum M$ is the same. However, in the case of axis B passing through one corner of the triangle, there are more masses farther away from the axis than in the case of axis A. Therefore, $R_B\gt R_A$
Because $\omega$ is the same for both cases, $KE_r$ is greater for axis B.
