(d) At height $h^{'},\frac{g'}{g} = 1 - \frac{2h}{R} = \frac{90}{100}$
or $\frac{2h}{R} = 1 - \frac{90}{100} = \frac{10}{100} = \frac{1}{10}$
or $R = 20h = 20 \times 320 = 6400\ km$
At dept $d,\frac{g'}{g} = 1 - \frac{d}{R} = \frac{95}{100}$
or $\frac{d}{R} = 1 - \frac{95}{100} = \frac{5}{100} = \frac{1}{20}$
or $d = \frac{R}{20} = \frac{6400}{20} = 320\ km$