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 \(67 \%\).