Let $F\subseteq \mathbb C$ be the splitting field of $x^7-2$ over $\mathbb Q$ and $z=e^{\frac{2\pi i}{7}}$ a primitive seventh root of unity. Let $[F:\mathbb Q(z)]=a$ and $[F:\mathbb Q(2^{\frac{1}{7}})]=b$. Then prove that $a>b$.
Now $F$ is the smallest field such that $x^7-2$ splits over $\mathbb Q$.
$x^7-2=(x-2^{\frac{1}{7}})(a_6x^6+a_5x^5+\cdots+a_0)=(x-2^{\frac{1}{7}})f(x)$
$x^7-2$ is an irreducible polynomial over $\mathbb Q$ by Eisenstein Criterion. Obviously $[F:\mathbb Q(2^{\frac{1}{7}})]=\deg (f(x))=6$.
Again $z $ satisfies $y^7-1=0\implies (y-1)(1+y+y^2+\cdots+y^6)=0$. Now $1+y+y^2+\cdots+y^6$ is an irreducible polynomial over $\mathbb Q$ whose one of the roots is $z$, so $1+y+y^2+\cdots+y^6=(y-z)g(x)$ where $\deg g(x)=5$ and $g(x)$ is irreducible over $\mathbb Q(2^{\frac{1}{7}})$.
Hence $b=5$.
Thus $a>b$.
Please help me to ensure whether it's correct or not.
You seem to have got mixed up a bit with your definitions of $a$ and $b$. You started off well by correctly calculating that $b=[F:\mathbb Q(\sqrt[7]2)]=6$. After that, you've done the wrong calculation: you should be calculating how $x^7-2$ splits in $\mathbb Q(z)$, not how $y^7-1$ splits in that field (it splits completely!)
There is an easier way to do this whole calculation. Observe that you can calculate $F$ explicitly: $$F = \mathbb Q(z, \sqrt[7]2).$$ Certainly $F$ must contain both $z$ and $\sqrt[7]2$, and $$X^7-2 = (X-\sqrt[7]2)(X-z\sqrt[7]2)\cdots(X-z^6\sqrt[7]2)$$splits completely in this field. You can then calculate $[F:\mathbb Q]$ using the coprimality of $6=[\mathbb Q(z):\mathbb Q]$ and $7=\mathbb [Q(\sqrt[7]2):\mathbb Q]$, and $a$ and $b$ can be found using the tower law.