Given data

$I_1 = 4\ \text{kg m}^2, \qquad I_2 = 2\ \text{kg m}^2$ $\omega_1 = 10\ \text{rad/s}, \qquad \omega_2 = 4\ \text{rad/s}$

When the two discs are brought into contact face-to-face with common axis, friction makes them rotate with a common final angular speed.

Use conservation of angular momentum

Since no external torque acts on the system,

$I_1\omega_1 + I_2\omega_2 = (I_1+I_2)\omega_f$

Substituting values:

$4(10) + 2(4) = (4+2)\omega_f$ $40 + 8 = 6\omega_f$ $48 = 6\omega_f$ $\omega_f = 8\ \text{rad/s}$

Initial kinetic energy

$K_i = \frac12 I_1\omega_1^2 + \frac12 I_2\omega_2^2$

$K_i = \frac12(4)(10^2) + \frac12(2)(4^2)$ $K_i = 2\cdot 100 + 1\cdot 16$ $K_i = 200 + 16 = 216\ \text{J}$

Final kinetic energy

$K_f = \frac12 (I_1+I_2)\omega_f^2$

$K_f = \frac12 (6)(8^2)$ $K_f = 3\cdot 64 = 192\ \text{J}$

Loss in kinetic energy

$\Delta K = K_i - K_f$ $\Delta K = 216 - 192 = 24\ \text{J}$

Final answer

The loss in kinetic energy is:

$\boxed{24}$