c
Frictional force on block $B$
$=\mu \mathrm{m}_{\mathrm{B}} \mathrm{g}=0.4 \times 3 \times 10=12 \mathrm{N}$
Hence, acceleration $=\frac{12}{3}=4 \mathrm{ms}^{-2}$
Hence, maximum force
$F=\left(\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}\right) \mathrm{a}=(6+3) \times 4=36 \mathrm{N}$