It is given that decoder is 3x8 with 2 enable inputs. Now, input lines must be split such that two of them act as enable and only 1 act as input line. As shown in below diagram :
Please note here that 1x8 DEMUX requires 3 select lines. Hence we need to use input line as 3rd select line.
Now we have 16 bits address having 2 bits for ENABLE inputs (A,B) and another bit C I will take for Input line.
So overall I am taking first three bits of address as DEMUX select lines so that it activates one of the 8 output lines at a time, which will select one of the 8 RAM blocks.
First two bits (AB) of the address will act as enable of decoder (because it is as mentioned in question), third bit C will act as input line, and combination of ABC together will activate the respective output line. A is MSB here and C will be LSB.
Now we have 13 bits that can be the maximum possible address bits which can be used for memory. So size of each memory block will be $2^{13}$ bits or $2^{10}$ Bytes or $1KB$
If you want to calculate the address Range :
Block 1 will get selected when ABC=000 and address range will be from 0000000000000000 to 0001111111111111 $[0000-1FFF]$ in Hex
Block 2 will get selected when ABC=001 and address range will be from 0010000000000000 to 0011111111111111 $[2000-3FFF]$ in Hex
Block 3 will get selected when ABC=010 and address range will be from 0100000000000000 to 0101111111111111 $[4000-5FFF]$ in Hex
And so on.... Hope its correct.