Answer
${\bf 6.28\times 10^{-5}}\;\rm Wb$, ${\bf 6.28\times 10^{-5}}\;\rm Wb$
Work Step by Step
We know that the magnetic flux inside the loop is given by
$$\Phi=AB\cos\theta$$
where $A$ is the cross-sectional area of the loop and $B$ is the external magnetic field which, in this case, is from the solenoid.
Let's assume that the solenoid is long enough that the magnetic field inside it is constant and outside it is zero.
This means that the part of the cross-sectional area of the loop that is affected by the solenoid's magnetic field is equal to the cross-sectional of the solenoid.
Hence,
$$\Phi=A_{\rm solenoid}B_{\rm solenoid}\cos\theta$$
When the loop is perpendicular to the solenoid, $\theta=0^\circ$ since the magnetic field direction inside the solenoid is along its length (The field of a solenoid is along the axis) and the area vector is across the loop center and both are parallel.
Thus,
$$\Phi=\pi r^2_{\rm solenoid}B_{\rm solenoid}\cos0^\circ$$
Plug the known;
$$\Phi=\pi (1\times 10^{-2})^2 (0.20)\cos0^\circ$$
$$\Phi_1=\color{red}{\bf 6.28\times 10^{-5}}\;\rm Wb$$
When the loop is tilted 60$^\circ$, the result will not change since the effective area of the loop increases while the magnetic field component in the direction of the loop area vector decreases. These two factors cancel each other and we get the same flux.
$$\Phi_2=\color{red}{\bf 6.28\times 10^{-5}}\;\rm Wb$$
