The value of $'g'$ at a certain height above the surface of the earth is $16$% of its value on the surface. The height is $(R = 6300 \ km)$

$\Rightarrow$ Value of $g^{\prime}$ at $h=16 \%$ of $g$ at Sugace

$\begin{array}{l} \Rightarrow g^{\prime}=\frac{16 g}{100}-\text { (1) } \\ \Rightarrow g^{\prime}=\frac{G M}{(R+h)^{2}} \\ \Rightarrow g^{\prime}=\frac{G M}{R^{2}\left(1+\frac{h}{R}\right)^{2}} \\ \Rightarrow g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}-\text { (11) }\left[\because \frac{G M}{R^{2}}=g\right] \end{array}$

(11) $\left[\because \frac{G M}{R^{2}}=g\right]$

From (1) and (II) we have

$\begin{aligned} & \frac{\not d}{\left(1+\frac{h}{R}\right)^{2}}=\frac{16 \%}{100} \\ \Rightarrow & \frac{100}{16}=\left(1+\frac{h}{R}\right)^{2} \\ \Rightarrow & \frac{10}{4}=1+\frac{h}{R} \\ \Rightarrow & \frac{6}{4}=\frac{h}{R} . \\ \Rightarrow & h=\frac{6}{4} R=\frac{3}{2} R . \\ & h=\frac{3}{2} \times 6300=9450 \mathrm{~km} \\ & h=9450 \mathrm{~km} \end{aligned}$

Hence, option (d) is correct