Answer
$\omega_A=32 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$\vec{v_B}=\vec{v_E}+\vec{\omega_A}\times \vec{r_{B/E}}~~$[eq(1)]
The velocity of $E$ is given as
$\vec{v_E}=\vec{\omega_{B}}\times \vec{r_E}$
$\implies \vec{v_E}=2\hat k\times (-20\hat j)=40\hat i$
The velocity from $E$ to $B$ is given as
$\vec{v_{B/E}}=5\hat j$
$\vec{v_B}=\vec{\omega_{BC}}\times \vec{r_{BC}}$
$\implies \vec{v_B}=-8\hat k\times (-15\hat j)$
$\implies \vec{v_B}=-120\hat i$
We plug in the known values in eq(1) to obtain:
$-120\hat i=40\hat i+(\omega_A)\hat \times 5\hat j$
$\implies -120\hat i=(40-5\omega_A)\hat i$
Equating $i$ components on both sides, we obtain:
$-120=40-5\omega_A$
This simplifies to:
$\omega_A=32 rad/s$
