Step 1: Set up the problem.
Let the three digits of the number be \(a\), \(b\), \(c\), each from \(0\) to \(9\), with \(a \neq 0\) since it is the leading digit of a three-digit number.
We are told \(a \times b \times c = 70\).
Step 2: Factor 70 using single digits.
Write \(70\) as a product of its prime factors:
\[ 70 = 2 \times 5 \times 7 \]
Each of \(2\), \(5\), \(7\) is already a single digit (\(1\) to \(9\)), so one natural triple of digits is \((2, 5, 7)\), since \(2 \times 5 \times 7 = 70\).
Step 3: Check there is no other valid triple of digits.
Any other way of splitting \(70\) into three whole-number factors would need to combine the primes \(2, 7, 5\) into groups, giving triples such as \((1, 7, 10)\), \((1, 2, 35)\), \((1, 5, 14)\), or \((1, 1, 70)\). In every one of these, one of the factors, \(10\), \(14\), \(35\), or \(70\), is not a valid single digit, since digits only go up to \(9\).
So \((2, 5, 7)\), in some order, is the only set of three digits whose product is \(70\).
Step 4: Find the sum.
\[ 2 + 5 + 7 = 14 \]
Step 5: Final Answer.
The sum of the digits of this three-digit number is 14, matching option (B).
\[ \boxed{14} \]