If the gravitational field in the space is given as - K r 2 . Taking the reference point to be at r = 2…

Given, $E=-\frac{K}{r^2}$

Gravitational field and gravitational potential is related as $E=-\frac{dV}{dr}$

So, $-\frac{dV}{dr}=\frac{-K}{r^2}$

Integrating the above relation,

${\int }_{10}^{V}dV={\int }_2^3\frac{K}{r^2}dr$

$\Rightarrow V-10=K\left[\frac{1}{2}-\frac{1}{3}\right]$

$$\begin{gathered} \Rightarrow V-10=\frac{K}{6} \\ \Rightarrow V=\frac{6}{6}+10 \\ \Rightarrow V=11 \mathrm{J} {\mathrm{kg}}^{-1} \end{gathered}$$