jmc

Moving everything to one side, we get $$2a^2+4b^2+c^2-4ab-2ac=0.$$We can write this equation as $$(a-2b{)}^2+(a-c{)}^2=0,$$so $b=\frac{a}{2}$ and $a=c.$ Then by the Law of Cosines, $$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+a^2-\frac{a^2}{4}}{2a^2}=\boxed{\frac{7}{8}}.$$