Answer
The mass of the pulley is $22kg$.
Work Step by Step
Let's call the tension in the rope pulling the 11-kg block $T_1$ and the tension in the rope pulling the 44-kg block $T_2$. Both point upward.
The 44-kg block falls downward with $a=1/2\times9.8=4.9m/s^2$. Therefore, $$mg-T_2=ma$$ $$T_2=m(g-a)=(44kg)(9.8-4.9)m/s^2=215.6N$$
The 11-kg block is pulled upward with similar acceleration $a=4.9m/s^2$. Therefore, $$T_1-mg=ma$$ $$T_1=m(g+a)=(11kg)(9.8+4.9)m/s^2=161.7N$$
The pulley's radius is denoted as $R$. Since there is rotation in the pulley, both $T_1$ and $T_2$ produce torques with lever arm $R$ $$\sum\tau=T_2R-T_1R=I\alpha$$
The pulley is considered a solid cylindrical disk, so $I=\frac{1}{2}M_{pulley}R^2$. We also have $\alpha=\frac{a}{R}$
$$T_2R-T_1R=\frac{1}{2}MR^2\times\frac{a}{R}=\frac{1}{2}MaR$$ $$T_2-T_1=\frac{1}{2}Ma$$ $$M=\frac{2(T_2-T_1)}{a}=22kg$$
