The sum of all three-digit natural numbers which are divisible by $7$ is

The correct option is C: $\left(70336\right)$

The smallest and the largest three-digit number, which are divisible by $7$ are $105$ and $994$ respectively.
So, the sequence of three-digit numbers which are divisible by $7$ is $105,112,119,\dots ,994$

Clearly, it is an A.P. with first term $a=105$ and common difference $d=7$.

Let there be ${}^{\prime }{n}^{\prime }$ terms in this sequence.
Then, $a_n=994$
$\Rightarrow a+(n-1)d=994\text{[Since}{a}_n=a+(n-1)d\text{]}\Rightarrow 105+(n-1)7=994\Rightarrow 105+7n-7=994\Rightarrow 7n+98=994\Rightarrow 7n=994-98\Rightarrow 7n=896\Rightarrow n=\frac{896}{7}\Rightarrow n=128$

Now, required sum
$S_n=\frac{n}{2}(a+l)$

$=\frac{128}{2}(105+994)$

$=64\left(1099\right)$

$=70336$

Hence, the sum of three-digit number divisible by $7$ is $70336$.