jmc

We know that $$\sqrt[3]{2}=a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\cdots }}}>a,$$and $$\sqrt[3]{2}=a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\cdots }}}<a+1.$$The integer $a$ that satisfies $a<\sqrt[3]{2}<a+1$ is $a=1.$

Then $$\sqrt[3]{2}-1=\frac{1}{b+\frac{1}{c+\frac{1}{d+\cdots }}},$$so $$\frac{1}{\sqrt[3]{2}-1}=b+\frac{1}{c+\frac{1}{d+\cdots }}.$$As before, $b$ must satisfy $$b<\frac{1}{\sqrt[3]{2}-1}<b+1.$$Rationalizing the denominator, we get $$\frac{1}{\sqrt[3]{2}-1}=\frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}=\frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{2-1}=\sqrt[3]{4}+\sqrt[3]{2}+1.$$We have that $$\sqrt[3]{4}+\sqrt[3]{2}+1>1+1+1=3.$$Also, $1.3^3=2.197>2$ and $1.6^3=4.096>4,$ so $$\sqrt[3]{4}+\sqrt[3]{2}+1<1.3+1.6+1=3.9<4,$$so $b=\boxed{3}.$