A unit positive point charge is taken slowly through an...

Since the electrostatic field is conservative, the work done in moving a charge depends only on the potential difference between the initial and final positions, regardless of the path taken.

The electric potential $V(r)$ at a distance $r$ ($r \ge R$) from the centre of a uniformly charged solid sphere of radius $R$ and charge density $\rho$ is given by:
$V(r) = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r}$

The total charge $Q$ of the sphere is:
$Q = \dfrac{4}{3}\pi R^3 \rho$

Substituting $Q$ into the potential formula, we get:
$V(r) = \dfrac{1}{4\pi\epsilon_0} \dfrac{\dfrac{4}{3}\pi R^3 \rho}{r} = \dfrac{\rho R^3}{3\epsilon_0 r}$

The initial position $A$ is at a distance $r_A = 2R$ from the centre. The potential at $A$ is:
$V_A = \dfrac{\rho R^3}{3\epsilon_0 (2R)} = \dfrac{\rho R^2}{6\epsilon_0}$

The final position $B$ is at a distance $r_B = 3R$ from the centre. The potential at $B$ is:
$V_B = \dfrac{\rho R^3}{3\epsilon_0 (3R)} = \dfrac{\rho R^2}{9\epsilon_0}$

The work done by the external agent in moving a unit positive charge ($q = 1$) slowly from $A$ to $B$ is equal to the change in its potential energy:
$W = q(V_B - V_A) = 1 \times \left( \dfrac{\rho R^2}{9\epsilon_0} - \dfrac{\rho R^2}{6\epsilon_0} \right)$

$W = \dfrac{\rho R^2}{\epsilon_0} \left( \dfrac{2 - 3}{18} \right) = -\dfrac{\rho R^2}{18\epsilon_0}$

The magnitude of the total work done is:
$|W| = \dfrac{\rho R^2}{18\epsilon_0}$

Comparing this with the given expression $\dfrac{\rho R^2}{n\epsilon_0}$, we find:
$n = 18$

Answer: $18$