Answer
The net torque required $\tau$ will be the greatest when the axis passes through B.
Work Step by Step
The triangle starts from rest, so $\omega_0=0$. It reaches its final velocity $\omega=10rad/s$ in $t=10s$. So the triangle's angular acceleration is $$\alpha=\frac{\omega-\omega_0}{t}=1rad/s^2$$
The torque required to bring the triangle to have $\alpha=1rad/s^2$ is $$\tau=I\alpha=\sum MR^2\alpha$$
The total mass in the triangle is always constant, no matter where the axis lies. But when the axis passes through B, there are more masses lying farther away from the axis than when the axis passes through A or C, meaning $R_B$ is the greatest out of 3.
Therefore, the net torque required $\tau$ will be the greatest when the axis passes through B.
