Answer
$\displaystyle \vec{E}(\frac{d}{2}) = -\frac{2V_0}{3d}$
Work Step by Step
We can find an equation for the electric field by taking the negative derivative of $V(x)$ since $\displaystyle \vec{E} = -\frac{dV}{d\vec{s}}$ where $\vec{s}$ is the position vector.
$\displaystyle V(x) = V_0\ln(1+\frac{x}{d})$
$\displaystyle \vec{E}(x) = -V'(x) = -V_0\frac{1}{1+\frac{x}{d}}(\frac{1}{d})$
$\displaystyle \vec{E}(x) =-V_0\frac{1}{\frac{d+x}{d}}(\frac{1}{d})$
$\displaystyle \vec{E}(x) =-V_0\frac{d}{d+x}(\frac{1}{d})$
$\displaystyle \vec{E}(x) =-\frac{V_0}{d+x}$
We are then asked to find the electric field halfway between the two electrodes, which is $x = \displaystyle \frac{d}{2}$
$\displaystyle \vec{E}(\frac{d}{2}) = -\frac{V_0}{d+\frac{d}{2}}$
$\displaystyle \vec{E}(\frac{d}{2}) = -\frac{V_0}{\frac{3d}{2}}$
$\displaystyle \vec{E}(\frac{d}{2}) = -\frac{2V_0}{3d}$
