Setup. An octahedron has $8$ triangular faces. Each face is defined by three adjacent vertices (donor atoms). We count faces whose three vertices contain exactly one N (from NH${}_3$) and two Cl.

cis-[Co(NH${}_3$)${}_4$Cl${}_2$]Cl. Donor set on the octahedron: $4$ N and $2$ Cl, with the two Cl on adjacent (cis) vertices. The two Cl atoms share an edge, and that edge is shared by exactly $2$ triangular faces. Each of those faces contains both Cl atoms plus one of the adjacent N atoms, so each is a (1 N, 2 Cl) face. The other $6$ faces have either three N atoms or one Cl with two N. Count: $2$ faces.

mer-[Co(NH${}_3$)${}_3$Cl${}_3$]. In the meridional isomer the three Cl atoms occupy three vertices forming a $T$-shape, i.e. one Cl at the centre with two Cl trans across it on the perpendicular axis. Pair the central Cl with each of the two outer Cls (these are cis-pairs sharing an edge). Each such Cl-Cl edge is shared by $2$ faces, and the third vertex of each such face is one of the three N atoms, giving a (1 N, 2 Cl) face. That gives $2\times 2=4$ faces. The two outer Cls are trans to each other and do not share an edge, so they generate no extra (1 N, 2 Cl) faces.

Total. $2+4=6$.