(b)

we know that

$g=\frac{G M}{R^2}$

The mass of the earth decrease by $25 \%$,

remaining $75 \%$ of

$M =\frac{75}{100}\,M$

radius of earth increase by $50 \%$,

new radius $=150 \%$ of $R=\frac{150}{100} R$

$g^{\prime}=\frac{G\left(\frac{75}{100} M\right)}{\left(\frac{150 R}{100}\right)^2}$

$g^{\prime}=G^M \frac{M}{R^2}\left[\frac{75}{100} \times \frac{10000}{150 \times 150}\right]$

$g^{\prime}=g \frac{5}{15}$

$g ^{\prime}=0.33\,g$

The acceleration due to gravity decreases by

$= g -0.33\,g$

$=0.67\,g$

The acceleration due to gravity decreases by

$=0.67 \times 100 \%$

$=67\,\%$

Hence the acceleration due to gravity at its surface decreases by nearly $67 \%$.