Given: Kinetic energy of the centre of mass $\mathrm{K}.\mathrm{E}=\frac{4}{3} \mathrm{Joule}$

The velocity of centre of mass $v_{cm}$ is equal to the summation of the momentum of each ball to the total mass of the system.

$v_{\mathrm{cm}}=\frac{2\times M+1\times 1}{M+1}=\frac{2 \mathrm{M}+1}{M+1}$

Now kinetic energy of the centre of mass is $\mathrm{K}.\mathrm{E}=\frac{1}{2}m{v}^2$ 

$\mathrm{K}. \mathrm{E} =\frac{1}{2} \left(M+1\right) {\left(\frac{2 \mathrm{M}+1}{M+1}\right)}^2=\frac{4}{3}$

${\left(2M+1\right)}^2=\frac{8(M+1)}{3}$

$3{\left(2M+1\right)}^2=8 \left(M+1\right)$

$12 {\mathrm{M}}^2+4 \mathrm{M}-5=0$

$M=\frac{1}{2}=0.5 \mathrm{kg}$