Page size = $2^{12}$
PAS = $2^{26}$
VAS = $2^{34}$
No of pages $= 2^{34}/2^{12} = 4M$
No of frames $= 2^{26}/2^{12} = 16K$
Page table entry has frame number, no of bits required to address frames $= log 2^{14} = 14 bits$
So Page Table size $= 4M * 14bits = 7MB$
Inverted Page table entry has frame number, no of inverted page table entry = no of frames.
So Page Table size $= 16k * 14bits = 28KB$