Given data

The gravitational field is $g(r)=-\frac{K}{r^2}$ with $K=6\ \text{J·cm/kg}$

Reference point: $r_1=2\ \text{cm}, \quad V(r_1)=10\ \text{J/kg}$

We need to find potential at $r_2=3\ \text{cm}$

Relation between gravitational field and potential

Gravitational field and potential are related by $g(r)=-\frac{dV}{dr}$

Since $g(r)=-\frac{K}{r^2}$ we get $-\frac{dV}{dr}=-\frac{K}{r^2}$ $\Rightarrow \frac{dV}{dr}=\frac{K}{r^2}$

Integrate to get potential difference

$V(r_2)-V(r_1)=\int_{r_1}^{r_2} \frac{K}{r^2}\,dr$

$=K\int_{2}^{3} r^{-2}\,dr$

$=K\left[-\frac{1}{r}\right]_{2}^{3}$

$=K\left(-\frac{1}{3}+\frac{1}{2}\right)$

$=K\left(\frac{1}{6}\right)$

Substitute $K=6\ \text{J·cm/kg}$

$V(3)-V(2)=6\times \frac{1}{6}=1\ \text{J/kg}$

Therefore, $V(3)=V(2)+1=10+1=11\ \text{J/kg}$

Final answer

$\boxed{11\ \text{J/kg}}$

So the correct option is B.

Check with stored answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.