The coefficient of static friction, μₛ, between block A of mass 2 kg and the table as shown in the…

For the blocks not to move, the tension T in the string must balance the weight of B, and the friction on A must balance T.
For block B: $T = m_Bg$
For block A, the maximum static friction available is $f_{max} = \mu_s N = \mu_s m_A g$.
For the maximum value of $m_B$ (blocks just on the verge of moving): $T = m_Bg = \mu_s(m_Ag)$
$\Rightarrow m_B = \mu_s m_A = 0.2 \times 2 = 0.4$ kg
So the maximum mass of block B is 0.4 kg.