The angular acceleration is related to the torque and moment of inertia by:
\[
\tau = I \alpha,
\]
where \(\tau = 2.56 \, \text{N.m}\) is the torque and \(I = \frac{1}{2} m r^2\) is the moment of inertia for a solid disc. Using \(m = 4 \, \text{kg}\) and \(r = 0.4 \, \text{m}\), we calculate the moment of inertia:
\[
I = \frac{1}{2} \times 4 \times (0.4)^2 = 0.32 \, \text{kg.m}^2.
\]
Then, the angular acceleration \(\alpha\) is:
\[
\alpha = \frac{\tau}{I} = \frac{2.56}{0.32} = 8 \, \text{rad/s}^2.
\]