Given angular position

$\theta(t)=5t^2-8t$

Find angular velocity and angular acceleration

Angular velocity: $\omega = \frac{d\theta}{dt} = 10t-8$

Angular acceleration: $\alpha = \frac{d\omega}{dt} = 10$

Evaluate at $t=2\,\text{s}$

$\omega(2)=10(2)-8=12\,\text{rad/s}$ $\alpha=10\,\text{rad/s}^2$

Moment of inertia of the disk

For a solid circular disk about an axis perpendicular to its plane through the center,

$I=\frac{1}{2}MR^2$

Torque applied

$\tau = I\alpha = \frac{1}{2}MR^2 \cdot 10 = 5MR^2$

Power delivered by torque

Instantaneous power in rotational motion is

$P=\tau\omega$

Therefore at $t=2$ s,

$P = 5MR^2 \times 12 = 60MR^2$

Match with options

$\boxed{60MR^2}$

So the correct option is A.