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(B) Since particle $C$ is moved without any acceleration,the change in kinetic energy is zero,$\Delta K.E. = 0$.By the work-energy theorem,the work do

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$\frac{2 G M m}{r}$

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(B) Since particle $C$ is moved without any acceleration,the change in kinetic energy is zero,$\Delta K.E. = 0$.By the work-energy theorem,the work done by an external agent plus the work done by gravitational forces is zero: $W_{	ext{ext}} + W_{	ext{grav}} = 0$.Therefore,$W_{	ext{ext}} = -W_{	ext{grav}} = \Delta U = U_f - U_i$.The gravitational potential energy of the system involving particle $C$ is $U = -\frac{GMm}{r_A} - \frac{GMm}{r_B}$,where $r_A$ and $r_B$ are the distances of $C$ from $A$ and $B$ respectively.Initial position: $C$ is at the midpoint,so $r_A = r/2$ and $r_B = r/2$.$U_i = -\frac{GMm}{r/2} - \frac{GMm}{r/2} = -\frac{2GMm}{r} - \frac{2GMm}{r} = -\frac{4GMm}{r}$.Final position: $C$ is at a point equidistant $r$ from $A$ and $B$,so $r_A = r$ and $r_B = r$.$U_f = -\frac{GMm}{r} - \frac{GMm}{r} = -\frac{2GMm}{r}$.Work done $W_{	ext{ext}} = U_f - U_i = -\frac{2GMm}{r} - (-\frac{4GMm}{r}) = \frac{2GMm}{r}$.

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