A

Step 1: Understand the configuration. We have a spherical shell of radius R with a uniform charge q, and a point charge q0 at its center. As the shell expands slowly from radius R to 2R, we need to find the work done by the electric forces.

Step 2: According to Gauss's Law, the electric field outside a uniformly charged spherical shell (at a distance greater than R) acts as if all the charge were concentrated at its center. Therefore, the electric field ( extbf{E}) at a distance r (where R < r < 2R) from the center is given by:
$$ E = \frac{k(q + q_0)}{r^2} $$ where k is Coulomb's constant.

Step 3: The work done (W) by electric forces during the expansion of the shell can be calculated as:
$$ W = \int_{R}^{2R} F \, dr $$
where F is the force acting on the shell due to the electric field. The force can be expressed as:
$$ F = q imes E = q \frac{k(q + q_0)}{r^2} $$

Step 4: Now, substituting this into the work integral:
$$ W = \int_{R}^{2R} q \frac{k(q + q_0)}{r^2} \, dr $$

Evaluating this integral:
$$ W = qk(q + q_0) \left[ -\frac{1}{r} \right]_{R}^{2R} = qk(q + q_0) \left( -\frac{1}{2R} + \frac{1}{R} \right) $$
$$ = qk(q + q_0) \left( \frac{1}{2R} \right) $$. Thus:
$$ W = \frac{qk(q + q_0)}{2R}. $$

Step 5: At this point, we notice that work done by an external agent against electric forces must equal the negative of the work done by the electric forces, hence:
$$ W_{external} = -W = -\frac{qk(q + q_0)}{2R}. $$

Final Answer: The work done by electric forces during the expansion is $$ \frac{qk(q + q_0)}{2R} $$ and the work done by an external agent against electric forces is $$ -\frac{qk(q + q_0)}{2R}. $$