Static Friction

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.

In the frame of the truck, a pseudo force $F = ma = 1 \times 5 = 5$ N acts on the block backward.

Limiting (maximum) static friction: $f_L = \mu_sN = \mu_smg = 0.6 \times 1 \times 10 = 6$ N
Since $F = 5$ N $\lt f_L = 6$ N, the block does not slip; it moves with the truck.
Static friction is self-adjusting and equals the force needed: $f_r = 5$ N
So the frictional force on the block is 5 N (it is this force that gives the block its 5 m/s$^2$ acceleration).