Which step would be the rate-determining step?

Well, the first step generates #"I"(g)#, which is used in the second step. #"I"_2(g)# is quite stable compared to #"I"(g)#, and needs light to become #"I"(g)#.

I would say that the second step is slow, but that it is due to not enough #"I"(g)# available as a result of not enough #"I"_2(g)# bonds cleaved.

But let's suppose it was just the second step, because it contains #"H"_2# as one of the reactants. Then the initially proposed rate law becomes:

#r(t) = k_2[I]^2[H_2]#

However, #"I"# is an intermediate and #[I]# must be replaced with #[I_2]#, since #"I"_2# is the other reactant. Now, step 1 is incorrectly written. It must be an equilibrium to agree with the proposed rate law...

#"I"_2(g) stackrel(k_1" ")(rightleftharpoons) 2"I"(g)#
#""^(" "" "" "k_(-1))#

#2"I"(g) + "H"_2(g) stackrel(k_2" ")(->) "H"_2"I"_2(g)#

#"H"_2"I"_2(g) stackrel(k_3" ")(->) 2"HI"(g)#

Then if we assume the fast equilibrium approximation, we get the equilibrium expression to be:

#r_1(t) = k_1[I_2] = r_(-1)(t) = k_(-1)[I]^2#

#=> K -= k_1/(k_(-1)) = ([I]^2)/([I_2])#

As a result,

#[I]^2 = k_1/(k_(-1))[I_2]#

And this gives:

#color(blue)(r(t) = (k_2k_1)/(k_(-1))[I_2][H_2])#

And this agrees with the proposed rate law, where #k = (k_2k_1)/(k_(-1))#.