jmc

We begin by multiplying the units digit: $3_4\times 3_4=9_{10}=21_4$. So, we write down a $1$ and carry-over the $2$. Moving on to the next digit, we need to evaluate $2_4\times 3_4+2_4=8_{10}=20_4$. Thus, the next digit is a $0$ and a $2$ is carried over. Finally, the leftmost digits are given by the operation $1_4\times 3_4+2_4=5_{10}=11_4$. Writing this out, we have \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & \stackrel{2}{1} & \stackrel{2}{2} & \stackrel{}{3}_4 \\ & & & \times & & & 3_4 \\ \cline{4-7} & & & 1 & 1 & 0 & 1_4 \\ \end{array}So our final answer is $\boxed{1101_4}$.