First of all, the Jacobian of the spherical coordinat transformation is $\rho^2\sin\phi$ not $\rho\sin\phi$.
Also, the upper bound for $\rho$ should have been $6\cos\phi$ not $3$. The equation $\rho = 3$ describes a sphere of radius $3$ centered at the origin while the equation $\rho = 6\cos\phi$ describes a sphere of radius $3$ centered at $(x,y,z) = (0,0,3)$.
After fixing those two errors, you should get an integral which evaluates to the correct answer.
Its almost correct! The only problem is the bounds for $\rho$. Since you want to describe
$\rho$ in terms of the angle $\phi$ you can proceed as follows:
Fix two angles $\theta\in [0,2\pi),\ $$\phi\in [\pi/6,5\pi/6]$. Then for this two fixed angles take the unique point $(\rho,\theta,\phi)\equiv (x,y,z)$ which lies in the cylinder $x^2+y^2=1$. You want to calculate the radius $\rho$ of this point. Using the relations of $(x,y,z)$ in terms of $(\rho,\theta,\phi)$ and the equation $x^2+y^2=1$ we get
\begin{align}
(\rho \cos\theta \sin\phi)^2&+(\rho \sin\theta \sin \phi)^2=1\\
&\implies \rho^2\sin^2\phi=1\\
&\implies \rho=\frac{1}{|sin\phi|}
\end{align}
and since $\phi\in [\pi/6,5\pi/6]$ we eventually end up with $\rho=1/\sin\phi=\csc \phi$. Now , from that point and since the cylinder is inside the sphere of radius $2$ you move till you hit the sphere (for fixed $\theta,\phi$). This will happen when $\rho$ becomes $2$. So, the bounds for $\rho$ must be $\csc \phi\leq \rho\leq 2$.
Hence, the volume is
$$\int_{0}^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc\phi}^{2}\rho^2\sin\phi\, d\rho d\phi d\theta$$
Best Answer
In spherical coordinates, the region can be written as $$ E=\{(\rho,\theta,\phi)\;|\; 0 \le \theta \le 2\pi , \frac{1}{\sqrt{2}\cos\phi}\le \rho \le 1,0 \le \phi\le \cos^{-1}\left( \frac{1}{\sqrt{2}}\right)(=\frac{\pi}{4})\} $$ It follows that the volume equals $$ \iiint_E \rho^2\sin \phi\; dV = \frac{2\pi}{3}-\frac{\pi}{\sqrt{2}}+\frac{\pi}{6\sqrt{2}} $$
Note.