The net force on block $B$ is equal to the static frictional force (the normal force and gravitational force cancel). So, the static frictional force equal
$\begin{align*}m_B \cdot a = \left(10 \mbox{ kg}\right) \cdot \left(2 \;\mathrm{m}\mathrm{/}\mathrm{s}^2 \right) = \boxed{20 \...$
Therefore, answer (A) is correct.

Solution to 2008 Problem 58