how to solve this one

Dear Student,
Maximum range of the voltmeter, V_max = 25 mV
So, in order to convert it into an ammeter of I_max = 25 A range, a shunt R is to be connected such that drop across R due to I_max = V_max, i.e.,
$$\begin{gathered} V_{max}=I_{max}R \\ R=\frac{V_{max}}{I_{max}}=\frac{25 mV}{25 A}=1 mohm = 0.001 ohms \end{gathered}$$

Regards