Solution

The sum of the forces acting on the volume of liquid highlighted in the figure is zero,so let's write force balance to the $x$ axis:

\begin{equation} \sigma L-\sigma L\sin\theta=P_{avg}Lh \end{equation}

where $P_{avg}$ is an average pressure of the liquid.As pressure linearly increasing with depth we get:

\begin{equation} P_{avg}=\frac{1}{2}\rho gh \end{equation}

Plugging eq(2) into eq(1) we get rise height of the liquid:

\begin{equation} h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}} \end{equation}

In the case of complete wetting, the contact angle is zero,so:

\begin{equation} h=\sqrt{\frac{2\sigma}{\rho g}} \end{equation}

For the water $\sigma=0.073 \frac{N}{m}$,$\rho=1000\frac{kg}{m^3}$ and $g=9.8\frac{m}{s^2}$,so:

\begin{equation} h \approx 3.9mm \end{equation}