Given,

$\mu_2=\frac{2}{3} \text { and } \mu_0=\frac{1}{3}$

Let the acceleration of the whole system be a then,

As per the diagram,

For block $B$,

$2 mg \cos 45^{\circ}+2 ma = T +\mu_0 N _{ b }$

$N _{ b }=2 mg \sin 45^{\circ}$

And, For block $A$,

$T =m g \cos 45^{\circ}+\mu_g N_a$

$N_{ a }=m g \sin 45^{\circ}$

So,

$T=m g \cos 45^{\circ}+\mu_2 m g \sin 45^{\circ} \ldots \ldots$

From eq $(1)$ and $(2)$

$2 m g \cos 45^{\circ}+2 ma =m g \cos 45^{\circ}+\mu_2 m g \sin 45^{\circ}+\mu_0 2 m g \sin 45^{\circ}$

After solving, We get $a =0$

So the acceleration of the block $A$ will be zero.