Given Information:
- LXI B, 2384H - This instruction loads the value 2384H into the register pair BC.
- LOOP: DCX B - This instruction decrements the register pair BC by 1.
- Clock frequency of MOV A, C is 2 MHz.
- ORA - This instruction performs a logical OR of the accumulator with the operand.
- JNZ LOOP - This instruction jumps to the label LOOP if the zero flag is not set.

Calculating the Delay:
To calculate the delay, we need to find the number of clock cycles required for each instruction and multiply it by the clock period.

1. LXI B, 2384H:
- This instruction requires 10 cycles to load a 16-bit value into BC.
- The clock period is the inverse of the clock frequency, which is 1/2 MHz = 0.5 microseconds.
- Therefore, the delay caused by this instruction is 10 * 0.5 microseconds = 5 microseconds.

2. LOOP: DCX B:
- This instruction requires 5 cycles to decrement the register pair BC by 1.
- The delay caused by this instruction is 5 * 0.5 microseconds = 2.5 microseconds.

3. MOV A, C:
- This instruction requires 5 cycles to move the value from register C to the accumulator.
- The delay caused by this instruction is 5 * 0.5 microseconds = 2.5 microseconds.

4. ORA:
- This instruction requires 4 cycles to perform a logical OR of the accumulator with the operand.
- The delay caused by this instruction is 4 * 0.5 microseconds = 2 microseconds.

5. JNZ LOOP:
- This instruction requires 10 cycles if the zero flag is not set, and 7 cycles if the zero flag is set.
- Since the zero flag is not set in this case, we consider the delay as 10 * 0.5 microseconds = 5 microseconds.

Total Delay:
The total delay is the sum of the delays caused by each instruction:
5 microseconds + 2.5 microseconds + 2.5 microseconds + 2 microseconds + 5 microseconds = 17 microseconds.

Therefore, the delay provided by the given set of instructions, neglecting the delay due to the LXI instruction, is 17 microseconds.