**Introduction:**
To find the smallest number by which the given numbers multiply to give a perfect cube of 7803, we need to factorize 7803 into prime factors and then identify the missing factors required to make it a perfect cube.

**Prime Factorization of 7803:**
To find the prime factors of 7803, we can begin by dividing it by the smallest prime number, which is 2. However, 7803 is an odd number and not divisible by 2. So, we move to the next prime number, which is 3. Dividing 7803 by 3, we get 2601. Continuing this process, we find that 2601 is divisible by 3, giving us 867. Dividing 867 by 3, we obtain 289. Again, dividing 289 by 17, we get 17.

Hence, the prime factorization of 7803 is 3 * 3 * 17 * 17.

**Identifying the Missing Factors:**
To make the product a perfect cube, we need each prime factor to have an exponent divisible by 3. In the given prime factorization, both 3 and 17 have an exponent of 2, which is not divisible by 3. Therefore, we need to find the missing factors to make the exponents divisible by 3.

**Factors Required to Make the Exponents Divisible by 3:**
For the prime factor 3, the exponent is 2, which is not divisible by 3. To make it divisible by 3, we need to multiply it by 3. So, the missing factor for 3 is 3.

For the prime factor 17, the exponent is also 2, which is not divisible by 3. Hence, we need to multiply it by 17 to make it divisible by 3. Therefore, the missing factor for 17 is 17.

**Calculating the Smallest Number Required:**
To find the smallest number by which the given numbers multiply to give a perfect cube of 7803, we multiply the missing factors we identified earlier.

Smallest number = Missing factor for 3 * Missing factor for 17
= 3 * 17
= 51

Hence, the smallest number by which the given numbers multiply to give a perfect cube of 7803 is 51.