1. Use the relation between deflection and current

For a moving coil galvanometer, deflection is directly proportional to the current through the galvanometer coil:

$\theta \propto I_g$

Initially, deflection is $25$ divisions. After applying the shunt, deflection becomes $5$ divisions.

So,

$\frac{I_g'}{I_g} = \frac{5}{25} = \frac{1}{5}$

Thus, after shunting, the current through the galvanometer becomes one-fifth of the original current.

2. Use current division in galvanometer and shunt

Let galvanometer resistance be $G$ and shunt resistance be

$S = 24\,\Omega$

When shunt is connected, the total current divides between galvanometer and shunt.

If $I_g'$ is current through galvanometer and $I_s$ through shunt, then since both are in parallel,

$I_g' G = I_s S$

So,

$\frac{I_s}{I_g'} = \frac{G}{S}$

Total current is

$I = I_g' + I_s = I_g'\left(1 + \frac{G}{S}\right)$

Originally, without shunt, the full current goes through galvanometer:

$I = I_g$

Hence,

$I_g = I_g'\left(1 + \frac{G}{S}\right)$

From step 1,

$I_g = 5 I_g'$

Therefore,

$5 = 1 + \frac{G}{S}$

$\frac{G}{S} = 4$

$G = 4S = 4 \times 24 = 96\,\Omega$

3. Final answer

$\boxed{G = 96\,\Omega}$

So the correct option is C.