The correct option of this question will be (c).
Solution —
Given $I_{s}^{\prime}=I_{s}+\frac{35}{100} I_{s}=\frac{135}{100} I_{s}$
Initial voltage sensitivity, $V_{s}=\frac{I_{s}}{R}$
New voltage sensitivity, $V_{s}^{\prime}=\frac{I_{s}^{\prime}}{R^{\prime}}$
$=\left(\frac{135}{100} I_{s}\right) \times \frac{1}{3 R}=\frac{9}{20} V_{s}$
% decrease in voltige sensitivity
$\left(\frac{V_{s}-V_{s}^{\prime}}{V_{s}}\right) \times 100 \%=\frac{V_{s}-\frac{9}{20} V_{s}}{V_{s}} \times 100 \%=55 \%$