Let the three children be $A,B$ and $C$ where $A$ is the first child

**Calculating the total number of case**

For each chocolate (assuming chocolates are distinguishable) we have $3$ choices, give it to $A$ or $B$ or to $C.$

So for ten chocolates we have total no. of choices $=\underbrace{3\times 3\times 3 \times \ldots \times 3}_{\text{ten times}}=3^{10}.$

**Calculating the number of favorable cases**

Firstly we have to select three chocolates out of ten that have to be given to $A$ and number of ways for this $={}^{10}C_3.$_{ }

For remaining $7$ chocolates, for each we have two choices i,e either to give it to $B$ or to $C.$

So, no of favorable cases $= {}^{10}C_3 \times 2^7.$

Our required probability $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$

$\quad \quad=\frac{{}^{10}C_3\times 2^7}{3^{10}} $

$\quad \quad =\frac{5\times 2^{10}}{3^9}$

i.e., option B