We have,
$a^2-c^2=b(b-c), \sqrt{2} a=2 b-c$ and $R=\frac{1}{\sqrt{3}}$
$\begin{aligned} & \Rightarrow \quad a^2-c^2=-b^2-b c \Rightarrow b^2+c^2-a^2=b c \\ & \Rightarrow \quad \frac{b^2+c^2-a^2}{2 b c}=\frac{1}{2} \Rightarrow \cos A=\frac{1}{2} \Rightarrow A=60^{\circ}\end{aligned}$
Since,
$\frac{a}{\sin A}=2 R \Rightarrow \frac{a}{\sin 60^{\circ}}=\frac{2}{\sqrt{3}} \Rightarrow a=1$
$\sqrt{2} a=2 b-c \Rightarrow 2 a^2=4 b^2+c^2-4 b c$
$\begin{aligned} & \Rightarrow \quad 2 a^2=4\left(b^2-b c\right)+c^2 \Rightarrow 2=4\left(a^2-c^2\right)+c^2 \\ & \Rightarrow \quad 2=4-3 c^2 \Rightarrow c^2=\frac{2}{3} \Rightarrow c=\frac{\sqrt{2}}{\sqrt{3}}\end{aligned}$
Now, $2 b=\sqrt{2} a+c$
$\Rightarrow \quad 2 b=\sqrt{2}+\frac{\sqrt{2}}{\sqrt{3}} \Rightarrow \quad b=\frac{\sqrt{2}(\sqrt{3}+1)}{2 \sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{6}}$