Step 1: Count the electron pairs around the central atom of water.
In $\text{H}_2\text{O}$, oxygen has $2$ bond pairs (to the two hydrogens) and $2$ lone pairs, giving it a bent, V shaped outline rather than a straight line.
Step 2: Do the same quick count for each option.
In $\text{CH}_4$ the central carbon has $4$ bond pairs and no lone pair, giving a tetrahedral shape. In $\text{CO}_2$ carbon has $2$ double bonds and no lone pair, giving a straight linear shape. In $\text{BeH}_2$ beryllium has $2$ bond pairs and no lone pair, again linear.
Step 3: Now check $\text{SO}_2$.
Sulfur here has $2$ bonding regions and $1$ lone pair sitting on it, and that lone pair pushes the bonds together just like the two lone pairs on oxygen bend the water molecule.
Step 4: Match the shapes.
Since $\text{SO}_2$ is the only option with a lone pair distorting it into a bent geometry, its overall shape resembles water far more than the straight or symmetrical shapes of the other three. \[ \boxed{\text{Structure of } \text{SO}_2 \text{ molecule}} \]