16 MB is program size and 2KB is frame/page size
Number of frames in Physical memory will be $\frac{2^{24}}{2^{11}}=2^{13}$
so we require $2^{13}$ entries in last level of page table
Possible breakup
12|11|11
13|10|11
14|9|11 and so on
let us take 12|11|11 so if we have only 11 bits for second level then we require $\frac{2^{13}}{2^{11}}=2^{2}$ 2nd level page table
let us take 13|10|11 so if we have only 10 bits for second level then we require $\frac{2^{13}}{2^{10}}=2^{3}$ 2nd level page table
let us take 14|9|11 so if we have only 9 bits for second level then we require $\frac{2^{13}}{2^{9}}=2^{4}$ 2nd level page table
so consider option 1
for 12|11|11 n=12 $\frac{2^{24}}{2^{34-12}}=2^{2}$ matches with our result
for 13|10|11 n=13 $\frac{2^{24}}{2^{34-13}}=2^{3}$ matches with our result
for 14|9|11 n=14 $\frac{2^{24}}{2^{34-14}}=2^{4}$ matches with our result
option 1 is correct