Identify the species in which the central atom undergoes sp^3d^2...

For $XeF_4$:
Central atom Xe has 8 valence electrons.
Number of bond pairs = 4
Number of lone pairs = $\dfrac{8 - 4}{2} = 2$
Steric number = $4 + 2 = 6$
Hybridisation is $sp^3d^2$ with 2 lone pairs.

For $BrF_5$:
Central atom Br has 7 valence electrons.
Number of bond pairs = 5
Number of lone pairs = $\dfrac{7 - 5}{2} = 1$
Steric number = $5 + 1 = 6$
Hybridisation is $sp^3d^2$ with exactly 1 lone pair.

For $PCl_5$:
Central atom P has 5 valence electrons.
Number of bond pairs = 5
Number of lone pairs = $\dfrac{5 - 5}{2} = 0$
Steric number = $5 + 0 = 5$
Hybridisation is $sp^3d$ with 0 lone pairs.

For $I_3^-$:
Central atom I has $7 + 1 = 8$ valence electrons.
Number of bond pairs = 2
Number of lone pairs = $\dfrac{8 - 2}{2} = 3$
Steric number = $2 + 3 = 5$
Hybridisation is $sp^3d$ with 3 lone pairs.

Thus, $BrF_5$ is the only species with $sp^3d^2$ hybridisation and exactly one lone pair.

Answer: $BrF_5$