A thin, metallic spherical shell contains a charge Q on it. A point...

According to the properties of conductors, the electric field inside the cavity of a conducting shell due to any external charges is always zero (electrostatic shielding).

The external charges in this case are the point charge $q_1$ and the charge distributed on the outer surface of the shell. Therefore, the electric field at the centre due to the outer surface charge ($\vec{E}_{\text{outer}}$) and the point charge $q_1$ ($\vec{E}_{q_1}$) must sum to zero:
$\vec{E}_{\text{outer}} + \vec{E}_{q_1} = 0 \Rightarrow \vec{E}_{\text{outer}} = -\vec{E}_{q_1}$

The positive charge $q$ at the centre induces a charge $-q$ on the inner surface of the shell. Since the shell is spherical and $q$ is exactly at the centre, this induced charge is uniformly distributed. The electric field at the centre due to this uniformly distributed inner surface charge is zero ($\vec{E}_{\text{inner}} = 0$).

The total electric field at the centre due to the shell is the sum of the fields due to its inner and outer surfaces:
$\vec{E}_{\text{shell}} = \vec{E}_{\text{inner}} + \vec{E}_{\text{outer}} = 0 - \vec{E}_{q_1} = -\vec{E}_{q_1}$

Since $q_1$ is a positive charge located to the right of the shell, its electric field $\vec{E}_{q_1}$ at the centre is directed towards the left. Consequently, the electric field due to the shell $\vec{E}_{\text{shell}}$ is directed towards the right.

Since the central charge $q$ is positive, the force exerted on it by the shell is $\vec{F}_{\text{shell}} = q\vec{E}_{\text{shell}}$, which is directed towards the right.