Number of mercury atoms in a drop of mercury
You can use dimensional analysis. It is a very useful tool to solve equations such as these. 

What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).

To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.

You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$

Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.

A: Mass of mercury:

$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{}
\frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)}
$$

Notice that this is dimensionally equivalent. Therefore, 

$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A}
$$

B: Atoms of mercury per drop:

$$\frac{atoms}{drop}=
\frac{N_A(atoms)}{(mol^{})}\cdot{}
\frac{(mol)}{Mr(g)}\cdot{}
\frac{1000(g)}{1(kg)}\cdot{}
\frac{\rho{(kg)}}{(m^3)}\cdot{}
\frac{(m^3)}{10^6(ml)}\cdot{}
\frac{V(ml)}{drop}$$

Notice that this is dimensionally equivalent. Therefore, 
$$\frac{atoms}{drop}=
\frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$