The correct order of bond angles in the following compounds/species is:

The bond angles of the given species can be determined using VSEPR theory and hybridisation principles described in the sources:

1. $CO_2$: The central carbon atom undergoes $sp$ hybridisation, resulting in a linear geometry with a bond angle of $180^\circ$ .
2. $NH_4^+$: The nitrogen atom is $sp^3$ hybridised and forms four bond pairs with no lone pairs. This results in a regular tetrahedral geometry with a bond angle of $109.5^\circ$ .
3. $NH_3$: The nitrogen atom is $sp^3$ hybridised but has three bond pairs and one lone pair. According to the sources, lone pair-bond pair (lp-bp) repulsion is greater than bond pair-bond pair (bp-bp) repulsion, which reduces the ideal tetrahedral angle to $107^\circ$ .
4. $H_2O$: The oxygen atom is $sp^3$ hybridised with two bond pairs and two lone pairs. Because lone pair-lone pair (lp-lp) repulsion is even stronger than lp-bp repulsion, the bond angle is further reduced to $104.5^\circ$ .

Comparing these values, the increasing order of bond angles is: $H_2O (104.5^\circ) < NH_3 (107^\circ) < NH_4^+ (109.5^\circ) < CO_2 (180^\circ)$.