Step 1: Recall capillary rise.
The height a liquid climbs inside a thin tube is \[ h = \frac{2T\cos\theta}{\rho g r} \] where $T$ is surface tension, $\theta$ is the contact angle, $\rho$ is density, $g$ is gravity, and $r$ is the tube radius.
Step 2: See the key link.
Everything except the radius stays the same here, since it is the same liquid. So the height depends only on the radius. \[ h \propto \frac{1}{r} \] A thinner tube lifts the liquid higher.
Step 3: Write the first case.
In the first tube of radius $r$, the rise is $4$ cm. \[ h_1 = \frac{2T\cos\theta}{\rho g r} = 4 \text{ cm} \]
Step 4: Write the second case.
The second tube has radius $r/2$. Putting this in, \[ h_2 = \frac{2T\cos\theta}{\rho g (r/2)} = 2\cdot\frac{2T\cos\theta}{\rho g r} = 2 h_1 \]
Step 5: Find the new height.
Since $h_1 = 4$ cm, \[ h_2 = 2\times 4 = 8 \text{ cm} \]
Step 6: State the answer.
Halving the radius doubles the rise. \[ \boxed{h_2 = 8 \text{ cm}} \]