The current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$$I_s = \frac{nBA}{C}$$

where $$n$$ is the number of turns in the coil, $$B$$ is the magnetic field, $$A$$ is the area of the coil, and $$C$$ is the torsional constant of the suspension wire.

We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $$n'$$, and the new area be $$A'$$.

Since the current sensitivity is increased by 25%, we have:

$$I_s' = 1.25I_s = \frac{n' B A'}{C}$$

The resistance ($R$) of the galvanometer coil is given by:

$$R = \rho \frac{l}{A}$$

where $$\rho$$ is the resistivity of the wire and $$l$$ is the length of the wire. Since the resistance is kept constant, we can write:

$$R' = \rho \frac{l'}{A'} = R$$

Now, the voltage sensitivity ($V_s$) of the galvanometer is given by:

$$V_s = I_s \times R$$

We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $$V_s'$$:

$$V_s' = I_s' \times R'$$

Since $$R = R'$$, we can write:

$$V_s' = 1.25I_s \times R = 1.25V_s$$

The percentage change in the voltage sensitivity is:

$$\frac{V_s' - V_s}{V_s} \times 100\% = \frac{1.25V_s - V_s}{V_s} \times 100\% = 25\%$$