Let the mass of the block $$B$$ be $$M.$$
In equilibrium, $$T - Mg = 0 \Rightarrow T = Mg\,.......\left( {\text{i}} \right)$$
If blocks do not move, then $$T = {f_s}$$
where, $${f_s}$$ = frictional force $$ = {\mu _s}R = {\mu _s}mg$$
$$\therefore T = {\mu _s}mg\,......\left( {{\text{ii}}} \right)$$
Thus, from Eqs. (i) and (ii), we have
$$\eqalign{
& Mg = {\mu _s}mg \cr
& {\text{or}}\,M = {\mu _s}m \cr
& {\text{Given,}}\,{\mu _s} = 0.2,\,m = 2\,kg \cr
& \therefore M = 0.2 \times 2 = 0.4\,kg \cr} $$