You can use the Random class:

int randomNumber = new Random().nextInt(9000) + 1000;

Create a random integer in range 0,10^k - 10^(k-1), and add 10^(k-1) - in your example. range 0,90000, and add 10000 to it.

Use the Random.nextInt(int) method for it.

Something like:

new Random().nextInt(90000) + 10000

private long generateRandomNumber(int n) {
    long min = (long) Math.pow(10, n - 1);
    return ThreadLocalRandom.current().nextLong(min, min * 10);
}

nextLong produces random numbers between lower bound inclusive and upper bound exclusive so calling it with parameters (1_000, 10_000) for example results in numbers 1000 to 9999. Old Random did not get those nice new features unfortunately. But there is basically no reason to continue to use it anyways.

There's a better way to get random numbers, and that's with java.util.Random. Math.random() returns a double (floating-point) value, but based on your request of a 3-digit number, I'm going to assume that what you really want is an integer. So here's what you do.

// initialize a Random object somewhere; you should only need one
Random random = new Random();

// generate a random integer from 0 to 899, then add 100
int x = random.nextInt(900) + 100;

You just need to format your random value using the expression

String.format("%04d", randomValue);

If randomValue is 762, the call to String.format returns "0762", as desired.

EDIT: In response to comment below

Please leave randomValue as an integer. You only want to use this code for user interaction i.e. formatting. So to print randomValue you would use

System.out.printf("%04d\n", randomValue);

and you should use String.format if there is any part of the program that needs to keep a string for user interaction, but there is no need to change the way you store randomValue.

To compare what the user types in to randomValue, you can use Integer.parseInt. So you can use

int testValue;
try {
    testValue = Integer.parseInt(inputLine, 10);
} catch (NumberFormatException ex) {
    // The user typed in something that isn't a number.  Alert the user somehow, and make
    // him or her try again
}

to get the typed-in value as an int, and just compare testValue and randomValue with ==.