Answer
$\oint_C D_n g ds=0$
Work Step by Step
Consider the First Green Identity: $\oint_C F \cdot n ds=\iint_D div F(x,y) dA$
When $\nabla^2 g=0$
This implies that $\oint_C \nabla g \cdot n ds=0$
Since, $F=\nabla g$
$\oint_C \nabla g \cdot n ds=\iint_D div (\nabla g) dA$
or, $\oint_C \nabla g \cdot n ds=\iint_D \nabla \cdot (\nabla g) dA=\iint_D \nabla^2 g dA=\iint_D (0) dA=0$
As per the statement, $D_ng$ is defined as $\nabla g \cdot n$
This implies that $\oint_C D_n g ds=0$ (proved)
Thus, the result is proved.
