Let galvanometer resistance and full-scale current be $G = \text{galvanometer resistance}, \qquad I_g = \text{full-scale current of galvanometer}$

Case 1: Galvanometer shunted by $5\,\Omega$ gives full-scale at total current $250\,\text{mA}$

Here the galvanometer and shunt are in parallel. At full-scale, $I = 250\,\text{mA} = 0.25\,\text{A}$ Current through galvanometer is $I_g$, so current through shunt is $I_s = 0.25 - I_g$

Since voltage across galvanometer and shunt is same, $I_g G = I_s \cdot 5$ $I_g G = 5(0.25 - I_g)$ $I_g G = 1.25 - 5I_g$ $I_g(G+5)=1.25 \qquad ...(1)$

Case 2: Galvanometer in series with $1050\,\Omega$ gives full-scale at $25\,\text{V}$

Now it acts as a voltmeter. At full-scale, current through galvanometer is $I_g$. Total series resistance is $G + 1050$ Hence, $25 = I_g(G+1050) \qquad ...(2)$

Use equation (1) to find $I_g$ relation

From (1), $I_g(G+5)=1.25$

From (2), $I_g(G+1050)=25$

Divide (2) by (1): $\frac{G+1050}{G+5} = \frac{25}{1.25} = 20$

So, $G+1050 = 20(G+5)$ $G+1050 = 20G + 100$ $950 = 19G$ $G = 50\,\Omega$

Final answer $\boxed{50}$