P is a point at a depth $\mathrm{d}=0.02{\mathrm{R}}_{\mathrm{E}}({\mathrm{R}}_{\mathrm{E}}\to \text{radius of Earth})$ and Q is a point at a height h, from the surface of Earth, where the g’ value is same. Then

$g(1-\frac{2h}{R_E})=g(1-\frac{d}{R_E})$

$\mathrm{h}=\frac{\mathrm{d}}{2}=\frac{0.02{\mathrm{R}}_{\mathrm{E}}}{2}=0.01{\mathrm{R}}_{\mathrm{E}}$

 Escape velocity at $\mathrm{Q}\Rightarrow \frac{1}{2}{\mathrm{mv}}_{\mathrm{Q}}^2$

$=\frac{\mathrm{GMm}}{({\mathrm{R}}_{\mathrm{E}}+0.01{\mathrm{R}}_{\mathrm{E}})}$

${\mathrm{v}}_{\mathrm{Q}}=\sqrt{\frac{2\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}}(1+0.01{)}^{-1/2}$

$={\mathrm{v}}_{\text{esc}}(1-0.005)=0.995{\mathrm{v}}_{\text{esc}}$

$\Rightarrow 0.5\mathrm{\%}$ less than surface value Potential at P :

${\mathrm{V}}_{\mathrm{P}}=-\frac{\mathrm{GM}}{2{\mathrm{R}}_{\mathrm{E}}^3}\left(3{\mathrm{R}}_{\mathrm{E}}^2-{0.98}^2{\mathrm{R}}_{\mathrm{E}}^2\right)$

$=-\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}\cdot \frac{1}{2}\left(3-(1-0.02{)}^2\right)$

$\simeq -\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}\cdot \frac{1}{2}(2+0.04)=-\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}(1+0.02)$

$\therefore \frac{1}{2}{\mathrm{mv}}_{\mathrm{p}}^2=\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}\mathrm{m}(1+0.02)$

$\Rightarrow {\mathrm{v}}_{\mathrm{p}}=\sqrt{\frac{2\mathrm{GM}}{{\mathrm{R}}_{\mathrm{E}}}}(1+0.02{)}^{1/2}$

${\mathrm{v}}_{\mathrm{esc}}(1+0.01)=1.01{\mathrm{v}}_{\mathrm{esc}}$

$\therefore {\mathrm{v}}_{\mathrm{p}}\to \text{ more than }{\mathrm{v}}_{\text{esc }}$

Clearly, the escape velocity from P will be more than that from surface as some energy is required to first bring it to the surface and then make it escape.