Our numbers are consecutive integers, so are c-2, c-1, c, c+1 and c+2. So we know 3c is a perfect square and 5c is a perfect cube. But in a perfect square, the exponents in your prime factorization are all even, so c will need to be divisible by at least one more 3. And in a perfect cube, your exponents in your prime factorization are all multiples of 3, so if 5c is a perfect cube, c itself will need to be divisible by at least 5^2.
Since we want the smallest value of c, we don't want c to be divisible by any unnecessary primes, so c = 3^a * 5^b. Since 3c is a perfect square, a+1 and b must be even, so a is odd and b is even. Since 5c is a perfect cube, the exponents in 3^a * 5^(b+1) must be multiples of 3, so a is divisible by 3, and b is one less than a multiple of 3. From all of that, the smallest positive value of a is 3, and the smallest positive value of b is 2, and c = (3^3)(5^2) = 675.
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