for Conventional Page table ->
no. of pages = virtual Memory size / page size
no. of pages = 2$^{34}$ B / 2$^{12}$ B ( assuming Memory is Byte Addressable)
= 2$^{22}$
Page table size = no. of pages * PTE size
= 2$^{22}$ * 2 B => 8 MB ( becoz frame no. bits = 14 bits ~ 2 Bytes)
for Inverted Page table ->
no. of frames = Physical memory size / page size
= 2$^{26}$ B / 2$^{12}$ B
= 2$^{14}$
Page table size = no. of frames * PTE size
= 2$^{14}$ * 3 B ( becoz page no bits = 22 bits ~ 3 Bytes)
= 48 KB