The most direct way to get at this problem, I think, is to realize that the work performed by an electric force on moving a charge from point to point is equal to qΔV, where q is the value of the charge being moved, and ΔV is the change in electric potential (in other words, voltage) between the two points.
When the charge is at infinity -- and considering this hemisphere of charge to be the only excess charge in space -- we can say the electric potential is zero (because the electric potential of this hemisphere goes to zero at infinity). So, the work done to move the charge from the center of the base to infinity is:
W = -q(Vf-Vo)= qVo
where Vo is the electric potential generated by the hemisphere at the center of its base. Now, we just have to calculate Vo.
Vo will be the sum of the contributions of every infinitesimal unit volume of charge at that position. I am going to use spherical coordinates to locate each infinitesimal volume unit on the sphere, with the center of the base of the hemisphere being the origin. In those terms, the volume of each unit is r2sinθdrdθdφ, where r is the radial distance to the charge, θ is its azimuthal angle (usually measured from the +x axis), and φ is its angle of elevation (usually measured from the xy plane). And each infinitesimal volume unit of charge is a distance r from the center of the base of the hemisphere.
The value of the charge in that infinitesimal volume unit is the product of the volume charge density t and the volume of that unit r2sinθdrdθdφ, so dq = t(r2sinθdrdθdφ).
Therefore, the contribution to the electric potential at the origin from a given unit volume of charge at (r,θ,φ) is:
dVo = kdq/d = kt(r2sinθdrdθdφ)/r = ktrsinθdrdθdφ
To get the total potential at the center of the hemisphere base, we have to integrate this dV expression over the entire hemisphere. This involves integrating the r direction from 0 to R, the θ direction for 0 to 2π and the φ direction from 0 to π (you could also switch the integration limits on θ and φ and get the same result). So, we have:
Vo = ∫0R ∫02π ∫0π ktrsinθdrdθdφ
k = the Coulomb constant, = 1/4πε (with ε = εo, if the dielectric constant is unity (=1))
I got Vo = -(t/4πε)(πR2) = -tR2/4ε
So, the work done by the electric field as the charge is moved from the base of the hemisphere to infinity is:
W = q(-tR2/4ε) = -qtR2/4ε
Note that this is the negative of what they want you to show. That is because this is the work done by the electric field. As the charge is moved from the center of the hemisphere base to infinity, there are two forces at work on it. The electric force of the hemisphere and the pulling force (actuating force) which is moving it away.
The work done by the electric force of the hemisphere on the charge is negative, as we showed above, This is because the hemisphere is positively charged, and the point charge q is negative. So, as the point charge is being pulled to infinity, the hemisphere is trying to pull it back. Therefore, the electric force of the hemisphere on the charge, as it is moved, is opposite the point charge's displacement. This makes for negative work, as we found.
Now, if we suppose the charge starts at rest at the center of the base of the hemisphere, and then ends at rest at infinity, the change in its kinetic energy overall must be zero (it starts and ends at rest). The work-energy theorem says that the net work done on an object must equal its change in kinetic energy. Hence, the work done by the two forces on the point charge as it is moved must add up to zero
Wpull+Welec = 0 --> Wpull = -Welec
So the work done by the pulling force must be equal in value and opposite in sign to the work done by the electric force during this move. So:
Wpull = qtR2/4ε
Hope this helps! If you have any questions about the process (and, without diagrams, I know it can be tricky), just let me know.