Information capacity of Ternary-based system over Binary-based
Binary (two states)

1 Megabyte = 1,048,576 bytes = 8,388,608 bits = 8,388,608 cells.  See https://en.wikipedia.org/wiki/Megabyte and https://en.wikipedia.org/wiki/Byte.  1 byte = 8 bits.

There are $2^{8,388,608}$ different possible values that can be stored in this much memory.

Ternary (three states)

Now a cell can have 3 possible values.  If we have 8,388,608 cells, each of which can have 3 possible values, then the entire memory can hold $3^{8,388,608}$ possible values.  This is the equivalent of $\lg 3^{8,388,608} = 8,388,608 \times \lg 3 \approx 8,388,608 \times 1.58 \approx 13,295,629$ bits.  This corresponds to $13,295,629/8 = 1,661,953$ bytes.

Therefore, this many ternary cells can store the same  amount of data as 1,661,953 of ordinary binary cells (as 1,661,953 MB of memory).