Step 1: Examine the geometry of $\text{SO}_2$.
Sulphur dioxide is $sp^2$ hybridized with one lone pair, giving a bent shape.
Step 2: State its bond angle.
Because of the lone pair and the double bond character, its O to S to O angle is about $119.5^\circ$.
Step 3: Examine the geometry of water.
Water is $sp^3$ hybridized with two lone pairs, also bent.
Step 4: State its bond angle.
The two lone pairs squeeze the H to O to H angle down to about $104.5^\circ$.
Step 5: Take the difference.
\[ \Delta\theta = 119.5^\circ - 104.5^\circ \]
Step 6: Evaluate.
This gives $15.0^\circ$, matching option 3.
\[ \boxed{15.0^\circ} \]