In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (g-acceleration due to gravity )

As shown below, a pulley and block system on a smooth double inclined plane

Equation of force for both blocks,

$\mathrm{mgcos}{30}^{\mathrm{o}}-\mathrm{T}=\mathrm{ma\_\_\_\_\_}\left(1\right)$

$\mathrm{T}-\mathrm{mgcos}{60}^{\mathrm{o}}=\mathrm{ma} \mathrm{\_\_\_\_\_}\left(2\right)$

from the equations,

$\mathrm{a}=\frac{\sqrt{3}-1}{4}\mathrm{g}$

${\mathrm{a}}_{\mathrm{CM}}=\left|\frac{{\mathrm{ma}}_1+{\mathrm{ma}}_2}{\mathrm{m}+\mathrm{m}}\right|$

${\mathrm{a}}_{\mathrm{CM}}=\left|\frac{\left(\frac{\sqrt{3}-1}{4}\right)\mathrm{g}\hat{\mathrm{i}}+\frac{\sqrt{3}-1}{4}\mathrm{g}\hat{\mathrm{j}}}{2}\right|$

${\mathrm{a}}_{\mathrm{CM}}=\frac{\mathrm{g}}{2}\sqrt{{\left(\frac{\sqrt{3}-1}{4}\right)}^2+{\left(\frac{\sqrt{3}-1}{4}\right)}^2}$

${\mathrm{a}}_{\mathrm{CM}}=\frac{\mathrm{g}}{2}\times \sqrt{2}\times \frac{\sqrt{3}-1}{4}=\left(\frac{\sqrt{3}-1}{4\sqrt{2}}\right)\mathrm{g}$

Hence the correct answer is $\left(\frac{\sqrt{3}-1}{4\sqrt{2}}\right)\mathrm{g}.$