Answer
See answers.
Work Step by Step
a. The current in the galvanometer that produces full-scale deflection is the reciprocal of the reported sensitivity.
$$I_{full-scale}=\frac{1}{35000\Omega/V}=2.9\times10^{-5}A$$
b. The resistance of the voltmeter is the full-scale voltage, 250V, multiplied by the reported sensitivity.
$$R=V_{full-scale}(sensitivity)=(250V)( 35000\Omega/V)=8.8\times10^6\Omega$$