jmc

We carry out the multiplication as we carry out multiplication in base $10$. Fortunately, we do not have to worry about carrying over, since we are only multiplying by digits of $0$ or $1$. Thus: \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & 1 & 0 & 1 & 1_2 \\ & & & \times & 1 & 0 & 1_2 \\ \cline{4-7} & & & 1 & 0 & 1 & 1_2 \\ & & 0 & 0 & 0 & 0 & 0_2 \\ + & 1 & 0 & 1 & 1 & 0 & 0_2 \\ \cline{1-7} & 1 & 1 & 0 & 1 & 1 & 1_2 \\ \end{array}When summing, we need to carry-over for the second digit from the left. Thus, the sum is equal to ${\boxed{110111}}_2$.