A simple circuit

First we convert the $R_2$ - $R_4$ - $R_6$ star connection to a delta connection ( Y- $\Delta$ transform ) as shown in the figure above. Then the three delta equivalent resistors are:

$\begin{aligned} R_{24} & = \frac{R_2R_4+R_4R_6+R_6R_2}{R_6} = \frac{2\times 4 + 4 \times 6 + 6 \times 2}{6} = \frac{44}{6} = \frac{22}{3} \Omega \\ R_{46} & = \frac{44}{2} = 22 \Omega \\ R_{62} & = \frac{44}{4} = 11 \Omega \end{aligned}$

From the equivalent circuit, we have:

$\begin{aligned} R_{eq} & = \left( 1 || 11 + 3 || \frac{22}{3} \right) || 22 || 5 \\ & = \left( \frac{11}{12} + \frac{66}{31} \right) || 22 || 5 \\ & = \frac{1133}{372} || 22 || 5 \\ & = \frac{1}{\frac{372}{1133} + \frac{1}{22} + \frac{1}{5}} \\ & = 1.742808799 \end{aligned}$

Therefore, $\left \lfloor 10^3R_{eq} \right \rfloor = \boxed{1742}$

Same as Chew-Seong Cheong . Checked by converting Star of 1, 2, 3 into delta.