Step 1: Identify what controls freezing point.
Depression in freezing point is a colligative property, so it depends only on the total number of dissolved particles, not on their nature. More particles means a lower freezing point.
Step 2: Write the working rule.
Effective particle count is given by \(i \times M\), where \(i\) is the van't Hoff factor (number of particles per formula unit) and \(M\) is the molarity. All solutions here are 0.1 M, so only \(i\) decides the answer.
Step 3: Find \(i\) for each solute.
Glucose: does not ionise, \(i = 1\). Urea: does not ionise, \(i = 1\). KCl: gives \(K^+ + Cl^-\), \(i = 2\). \(CaCl_2\): gives \(Ca^{2+} + 2Cl^-\), \(i = 3\).
Step 4: Compare the effective particle counts.
\(CaCl_2\) gives the largest value \(0.1 \times 3 = 0.3\), which is more than KCl (0.2) and glucose or urea (0.1).
Step 5: Conclude.
The most particles cause the greatest lowering, so 0.1 M \(CaCl_2\) has the lowest freezing point.
\[ \boxed{0.1\ M\ CaCl_2} \]