Prepare the W state and measure its entanglement

Starting from $\lvert000\rangle$, build a circuit that outputs $\lvert W_3\rangle=\frac1{\sqrt3}\big(\lvert001\rangle+\lvert010\rangle+\lvert100\rangle\big).$ Qubit $0$ is the leftmost (most significant) bit, $R_Y(\alpha)=e^{-i\alpha Y/2}$, so $R_Y(\alpha)\lvert0\rangle=\cos\frac\alpha2\lvert0\rangle+\sin\frac\alpha2\lvert1\rangle$, and $\mathrm{CRY}(c,t,\alpha)$ applies $R_Y(\alpha)$ to qubit $t$ when qubit $c$ is $\lvert1\rangle$. Allowed gates: $R_Y$, $X$, $\mathrm{CRY}$ and $\mathrm{CNOT}$, at most $8$ gates in total. The grader requires the fidelity $F=\lvert\langle W_3\vert\psi\rangle\rvert^2\ge1-10^{-6}$.

A workable plan, following the sequential construction of the lesson: (1) put the amplitudes $\sqrt{2/3}$ and $\sqrt{1/3}$ on qubit $0$; (2) split the weight-$\tfrac23$ branch evenly with one controlled rotation onto qubit $1$, so that qubits $0,1$ hold $\tfrac1{\sqrt3}(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle)$; (3) write the missing bit into qubit $2$.

Let $\rho_0=\mathrm{Tr}_{12}\lvert W_3\rangle\langle W_3\rvert$ be the reduced state of qubit $0$. Compute its von Neumann entropy $S(\rho_0)=-\mathrm{Tr}\,\rho_0\log_2\rho_0=-\sum_\alpha\lambda_\alpha^2\log_2\lambda_\alpha^2,$ where $\lambda_\alpha$ are the Schmidt coefficients of $\lvert W_3\rangle$ across the cut between qubit $0$ and qubits $1,2$. Give the value in bits ($\log_2$). The grader accepts an absolute error of $10^{-3}$.