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The two-digit number which gives a remainder of $3$ when divided by $7$ are : $10, 17, 24, \dots, 94.$

Let $S = 10 + 17 + 24 + \dots + 94$

We know that, the sum of arithmetic progression: $S = \frac{n}{2}[$ First term $ + $ Last term $]$, where $n =$ number of terms.

Also, $n^{\text{th}}$ term of the AP, $T_{n} = a + (n-1)d$, where $a =$ first term, $d =$ common difference.

$\Rightarrow 94 = 10 + (n-1)7$

$\Rightarrow 94 = 10 + 7n-7$

$\Rightarrow 3n + 3 = 94$

$\Rightarrow 7n = 91$

$\Rightarrow \boxed{n = 13}$

Now, $S = \frac{13}{2}[10 + 94] = \frac{13}{2} \times 104 = 13 \times 52$

$\Rightarrow \boxed{S = 676}$

$\textbf{Short Method:}$ We can write all such numbers are of the form: $7k + 3$

- The smallest value of $k = 1$
- The largest value of $k = 13$ (Such that it is still a two-digit number)

Now, sum of all such numbers $ = 7(1) + 3 + 7(2) + 3 + 7(3) + 3 + \dots + 7(13) + 3$

$\qquad \qquad = 7(1 + 2 + 3 + \dots + 13) + 3 \times 13$

$\qquad \qquad = \frac{7 \times 13 \times 14}{2} + 39 = 91 \times 7 + 39 = 637 + 39 = 676$.

Correct Answer $:\text{B}$

$\textbf{PS:}$

- $1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$
- $1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n(n + 1)(2n + 1)}{6}$
- $1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = \left [\frac{n(n + 1)}{2}\right]^{2}$