For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div / mA will be:

We are told that the galvanometer shows a total deflection of 50 divisions when a potential difference of 50 mV is applied across its terminals.

Current sensitivity is given as 2 divisions per milliampere. By definition,

$$S_i=\frac{\text{deflection in divisions}}{\text{current in ampere}}.$$

Re-arranging, the current needed for any given deflection is

$$I=\frac{\text{deflection}}{S_i}.$$

For the full-scale deflection of 50 divisions we therefore have

$$I=\frac{50\ \text{div}}{2\ \text{div}\!/\!\text{mA}}.$$

Divisions cancel, leaving

$$I=25\ \text{mA}.$$

Now we already know that the corresponding voltage is 50 mV, i.e.

$$V=50\ \text{mV}=50\times10^{-3}\ \text{V}.$$

Ohm’s law states $$V=IR,$$ so the resistance of the galvanometer is

$$R=\frac{V}{I}=\frac{50\times10^{-3}\ \text{V}}{25\times10^{-3}\ \text{A}}.$$

Both the $$10^{-3}$$ factors cancel, giving

$$R=\frac{50}{25}=2\ \Omega.$$

Hence, the correct answer is Option B.