Answer
$I_B=4.4kg.m^2$
Work Step by Step
Before the disks are linked together, we have $$L_0=L_A+L_B=I_A\omega_A+I_B\omega_B$$
After the disks are linked together, $$L=(I_A+I_B)\omega=I_A\omega+I_B\omega$$
Angular momentum is conserved, so $$I_A\omega_A+I_B\omega_B=I_A\omega+I_B\omega$$ $$I_B(\omega-\omega_B)=I_A(\omega_A-\omega)$$ $$I_B=\frac{I_A(\omega_A-\omega)}{\omega-\omega_B}$$
We have $I_A=3.4kg.m^2$, disk A's initial velocity $\omega_A=+7.2rad/s$, disk B's initial velocity $\omega_B=-9.8rad/s$ and the 2 disks' final velocity $\omega=-2.4rad/s$
Therefore, $I_B=4.4kg.m^2$
